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. $N = 250\text{ rpm} = 4.167\text{ rev/s}$; ISO VG 100 oil; $L = 1.2D$; no-load loss limit $P_f \le 2.5\times10^{-4}\text{ hp} = 0.186\text{ W}$; diametral clearance $c_d = 0.0045D$ (radial $c_r = 0.00225D$).

Find. the maximum journal diameter $D$ and the allowable (oil) temperature limit.

Approach. The lightly-loaded (concentric) friction torque is given by Petroff’s equation; expressing it with $L$ and $c_r$ proportional to $D$ collapses the power loss to the form $P_f = K\,\mu D^3$. The budget fixes the product $\mu D^3$; the oil viscosity is then set by its temperature through the Walther (ASTM D341) chart, so the largest diameter corresponds to the lowest safe viscosity, i.e. the highest allowable oil temperature.

1. Petroff friction torque and power. For a concentric journal, $T = \dfrac{4\pi^2\mu\,r^3 L\,N}{c_r}$ and the loss is $P_f = T(2\pi N)$. Substituting $r=D/2$, $L=1.2D$, $c_r=0.00225D$: $$P_f = \frac{8\pi^3 \mu N^2}{c_r}\,r^3 L = 2.87\times10^{5}\,\mu\,D^3 \quad(\text{SI: W, Pa}\cdot\text{s, m}).$$
2. Impose the power budget. $$\mu D^3 = \frac{P_f}{2.87\times10^{5}} = \frac{0.186}{2.87\times10^{5}} = 6.49\times10^{-7}\ \text{Pa}\cdot\text{s}\cdot\text{m}^3.$$ Larger $D$ demands lower $\mu$, i.e. hotter, thinner oil.
3. Oil viscosity vs temperature (Walther). For ISO VG 100, $\nu = 100\text{ cSt}$ at $40^\circ\text{C}$ and $11.4\text{ cSt}$ at $100^\circ\text{C}$. Taking the practical continuous-service limit for mineral oil, $T_{max}\approx 70^\circ\text{C}$, the chart gives $\nu\approx 27.7\text{ cSt}$ and (with $\rho\approx 854\text{ kg/m}^3$) $\mu \approx 0.0236\ \text{Pa}\cdot\text{s}$.
4. Maximum diameter. $$D = \left(\frac{6.49\times10^{-7}}{0.0236}\right)^{1/3}.$$ $$\boxed{D_{max} \approx 30\text{ mm}, \qquad T_{allow} \approx 70^\circ\text{C}.}$$ Running cooler (higher $\mu$) would exceed the loss budget at this diameter, so $70^\circ\text{C}$ is the binding oil temperature; the same $\mu D^3$ line gives $\approx 20\text{ mm}$ at $40^\circ\text{C}$ and $\approx 41\text{ mm}$ at $100^\circ\text{C}$.