---
title: "Petroff equation for bearing friction"
subject: "Machine Design I"
chapter: 7
key: "semester_6/machine_design_i:petroff-equation-bearing-friction"
source: https://www.bctengineer.com/q/semester_6/machine_design_i/petroff-equation-bearing-friction
site: BCT Engineering
---

# Petroff equation for bearing friction

Machine Design I · Chapter 7: Lubrication and journal bearings

- Asked in: year not given, 6 marks
- Video: [Petroff's Equation](https://www.youtube.com/watch?v=1ZHor_tdLuk) — Mech Entertainment ⚙️
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/machine_design_i/petroff-equation-bearing-friction
- Whole chapter: https://www.bctengineer.com/md/semester_6/machine_design_i/chapter-7

## Question

State Petroff's law and derive the expression for the coefficient of friction of a lightly loaded journal bearing. A journal of diameter 75 mm and length 75 mm runs at 1000 rpm in a bearing with radial clearance 0.075 mm. The lubricant viscosity is 0.025 Pa.s and the radial load is 3 kN. Find the coefficient of friction, friction torque, power loss and the Sommerfeld number.

## Short answer

Petroff's law derives from shear stress μU/c, giving f = 2π² (μN/p)(r/c). For r=37.5mm, c=0.075mm, r/c=500, p=0.533 MPa, μN/p=7.81e-7, f=0.00771; torque T=0.867 N·m; power loss=90.8 W; Sommerfeld number S=0.195.

## Answer

#### Petroff's law

Petroff assumed a lightly loaded journal running concentric with the bearing, with the clearance $c$ filled with oil, so that the shear of the oil film gives the friction. The law states that the friction torque is proportional to viscosity and speed and inversely proportional to the clearance.

#### Derivation

Journal radius $r$, length $l$, speed $N$ rev/s, load $W$.

Surface velocity $U = 2\pi r N$. Shear stress in the film $\tau = \mu U/c = \dfrac{2\pi\mu r N}{c}$.

Friction force $F = \tau \times (2\pi r l)$ and torque $T = F\,r$:

$$
T = \frac{4\pi^2 \mu r^3 l N}{c}
$$

Also $T = f W r$ and $W = p\,(2r)\,l$ (projected area pressure). Therefore

$$
f = \frac{T}{W r} = \frac{4\pi^2\mu r^3 l N}{c \cdot p\,(2rl)\, r} = 2\pi^2\,\frac{\mu N}{p}\,\frac{r}{c}
$$

This is the Petroff equation.

#### Numerical

Data: $r = 37.5$ mm, $l = 75$ mm, $c = 0.075$ mm, $r/c = 500$, $N = 1000/60 = 16.67$ rev/s, $\mu = 0.025$ Pa.s, $W = 3000$ N.

$$
p = \frac{W}{ld} = \frac{3000}{0.075 \times 0.075} = 0.533\times10^6\ \text{Pa}
$$

$$
\frac{\mu N}{p} = \frac{0.025 \times 16.67}{0.533\times10^6} = 7.81\times10^{-7}
$$

$$
f = 2\pi^2 \times 7.81\times10^{-7} \times 500 = 0.00771
$$

$$
\begin{aligned} T &= f W r = 0.00771 \times 3000 \times 0.0375 = 0.867\ \text{N.m} \\ \text{Power loss} &= T\omega = 0.867 \times 2\pi \times 16.67 = 90.8\ \text{W} \end{aligned}
$$

$$
S = \left(\frac{r}{c}\right)^2\frac{\mu N}{p} = 500^2 \times 7.81\times10^{-7} = 0.195
$$

**Answer: f = 0.0077; T = 0.867 N.m; power loss = 90.8 W; Sommerfeld number = 0.195.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What does Petroff assume in deriving the friction in a lightly loaded bearing?
   A. The journal is concentric and the clearance is full of oil sheared like a Newtonian film, tau = mu U/c
   B. The journal is heavily eccentric with a thin film
   C. The surfaces touch at asperities
   D. The oil is supplied by an external pump
   Answer: A. Petroff took a concentric journal, with shear stress in the clearance tau = mu U/c giving the friction torque. (Key point: Petroff assumption)

2. Which is the Petroff expression for the coefficient of friction?
   A. f = 2 pi^2 (mu N / p)(r/c)
   B. f = 4 pi^2 mu r^3 l N / c
   C. f = (r/c)^2 mu N / p
   D. f = pi (mu N / p)(c/r)
   Answer: A. Dividing T = 4 pi^2 mu r^3 l N / c by W r with W = p(2r)l gives f = 2 pi^2 (mu N/p)(r/c). The second option is the torque. (Key point: Petroff equation)

3. For the 75 mm journal (r/c = 500, mu N/p = 7.81e-7), what is f?
   A. 0.0039
   B. 0.0077
   C. 0.0154
   D. 0.195
   Answer: B. f = 2 pi^2 x 7.81e-7 x 500 = 0.0077. The value 0.195 is the Sommerfeld number. (Key point: Numerical result)

4. How is the power loss found after the friction torque T = 0.867 N.m is known?
   A. P = T x N in rpm
   B. P = T x omega = T x 2 pi N (rev/s), about 90.8 W
   C. P = f x W
   D. P = T / omega
   Answer: B. Power equals torque times angular speed, 0.867 x 2 pi x 16.67 = 90.8 W. (Key point: Power loss)

## Related in Machine Design I

- [Journal bearing design with film thickness check](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Ajournal-bearing-minimum-film-thickness) — Machine Design I, chapter 7
- [Types of lubrication and viscosity](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Alubrication-types-viscosity) — Machine Design I, chapter 7
- [ASME code equation for shaft diameter](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Aasme-code-shaft-diameter-derivation) — Machine Design I, chapter 5

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- [Torque to overcome bearing friction](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Ajournal-bearing-torque) — Fluid Mechanics, chapter 1
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