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25-Nav-B1 Applied Thermodynamics and Heat Transfer (25-Mec-A1) · May 2013

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

National Examinations, May 2013 — 98-Mar-B1 Advanced Machine Design, 3 hours, open book (Part I: Questions 1–2 compulsory; Part II: any THREE of Questions 3–6; all six answered below for full study coverage).

Reference texts: R. G. Budynas & J. K. Nisbett, Shigley’s Mechanical Engineering Design, 10th ed. (shafts §7, bolted joints §8, bearings §12, brakes §16); R. L. Norton, Machine Design: An Integrated Approach, 5th ed.; R. C. Hibbeler, Mechanics of Materials, 10th ed. (impact loading, beam deflection); R. C. Juvinall & K. M. Marshek, Fundamentals of Machine Component Design, 5th ed. (journal bearings, friction brakes).

Check: this paper, although listed under Applied Thermodynamics and Heat Transfer, is headed “98-Mar-B1” with printed title “Advanced Machine Design”; it is a general mechanical machine-design paper — stress/yield criteria, journal bearings, impact loading, shaft fatigue, bolted-joint preload, and drum brakes — with zero thermodynamics or heat-transfer content, and it is solved as the exam actually printed.

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.

ω = 250 rpm bearing journal cₕ = 0.00225 D L = 1.2 D
Concentric (no-load) journal: uniform film $c_r$; Petroff shear over the whole bore sets the friction torque.
  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 $100^\circ\text{C}$.
Question 4 — results (loss budget $\mu D^3 = 6.49\times10^{-7}$)
Oil temperature$\mu$ (Pa·s)Max diameter $D$
$40^\circ\text{C}$$0.087$$20\text{ mm}$
$70^\circ\text{C}$ (design limit)$0.024$$\approx 30\text{ mm}$
$100^\circ\text{C}$$0.0095$$41\text{ mm}$

Check / assumptions: “No-load power loss” is the concentric Petroff loss, which fixes only the product $\mu D^3$; a second constraint is needed to isolate $D$. The maximum diameter is therefore taken at the highest sound operating temperature for a mineral oil ($\approx 70^\circ\text{C}$, above which oxidation life falls sharply), where $\mu$ is smallest. Density and viscosity–temperature values are from the standard ISO VG 100 Walther fit; a different assumed thermal limit shifts $D_{max}$ along the tabulated line.