---
title: "Hydrodynamic lubrication and Petroff's equation"
subject: "Internal Combustion Engines"
chapter: 6
key: "semester_6/internal_combustion_engines:hydrodynamic-lubrication-petroff-numerical"
source: https://www.bctengineer.com/q/semester_6/internal_combustion_engines/hydrodynamic-lubrication-petroff-numerical
site: BCT Engineering
---

# Hydrodynamic lubrication and Petroff's equation

Internal Combustion Engines · Chapter 6: Engine lubrication systems

- Asked in: year not given, 3+5 marks
- Video: [Derivation of petroff's equation || Engineering methods || About bearing and shaft](https://www.youtube.com/watch?v=31HGeVehgfI) — G Tech Education
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/internal_combustion_engines/hydrodynamic-lubrication-petroff-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_6/internal_combustion_engines/chapter-6

## Question

Explain the hydrodynamic theory of lubrication in a journal bearing. A lightly loaded journal bearing has shaft diameter 100 mm, bearing length 80 mm and radial clearance 0.05 mm. It carries a radial load of 4000 N at 1500 rpm. The oil viscosity is 0.025 Pa s. Using Petroff's equation, find the coefficient of friction, the friction torque and the power lost.

## Short answer

Hydrodynamic lubrication occurs when the rotating shaft drags oil into a converging wedge, building pressure that lifts the shaft onto a full fluid film, so surfaces are separated and friction is only viscous shear. Using Petroff's equation f = 2π² (μN/P)(r/c) with N=25 rev/s, bearing pressure P=5×10⁵ Pa, r/c=1000 yields f=0.0247; friction torque T=fWr=4.93 N·m; power loss=2πNT=0.775 kW.

## Answer

#### Hydrodynamic lubrication
When the shaft is at rest it touches the bearing at the bottom. As it begins to rotate, the shaft drags oil into the converging wedge between itself and the bearing because of the viscosity. At sufficient speed the oil in the wedge builds a pressure that lifts the shaft on a full fluid film, with the shaft centre displaced from the bearing centre (eccentricity $e$). The surfaces no longer touch, and the friction is only the viscous shear of the oil.

```text
   bearing
  +-----------+
  |  .--O--.  |   O = journal centre (displaced by e)
  | /  shaft\ |   thin film thickest on the entry side,
  | \_______/ |   thinnest at h_min on the exit side
  +-----------+   pressure peak ahead of h_min
```

The film thickness and load capacity depend on the **Sommerfeld number** $S = (\mu N/P)(r/c)^2$, where $\mu$ is the viscosity, $N$ the speed in rev/s, $P = W/(dL)$ the bearing pressure, $r$ the journal radius and $c$ the radial clearance. Higher viscosity and speed and lower load increase the film thickness. The friction coefficient follows the Stribeck curve: high at start (boundary), minimum when the film forms, then slowly rising in the fluid regime.

#### Petroff's equation
For a concentric journal (light load) the shear stress is $\tau = \mu U/c$ and the friction coefficient is

$$
f = 2\pi^2 \frac{\mu N}{P} \frac{r}{c}
$$

#### Calculation
$$
\begin{aligned}
N &= \frac{1500}{60} = 25\ \text{rev/s} \\
P &= \frac{W}{dL} = \frac{4000}{0.1 \times 0.08} = 5 \times 10^5\ \text{Pa} \\
\frac{r}{c} &= \frac{0.05}{0.00005} = 1000 \\
f &= 2\pi^2 \times \frac{0.025 \times 25}{5 \times 10^5} \times 1000 = 0.02467
\end{aligned}
$$

$$
\begin{aligned}
T &= f W r = 0.02467 \times 4000 \times 0.05 = 4.93\ \text{N m} \\
\text{Power loss} &= 2\pi N T = 2\pi \times 25 \times 4.93 = 775\ \text{W}
\end{aligned}
$$

**Answer:** $f = 0.0247$, friction torque $= 4.93$ N m, power loss $\approx 0.775$ kW.

## Check yourself

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

1. How does a full fluid film form in a hydrodynamic journal bearing?
   A. Oil is pumped in at high pressure from outside
   B. The rotating shaft drags oil into a converging wedge, building pressure that lifts the shaft
   C. The shaft is heated until oil boils
   D. The shaft touches the bearing at the top
   Answer: B. Viscous drag pulls oil into the converging wedge; at sufficient speed the pressure lifts the shaft so surfaces do not touch. (Key point: How the oil film forms)

2. How does the film thickness change with viscosity, speed and load according to the Sommerfeld number?
   A. Higher viscosity and speed and lower load increase it
   B. Higher load increases it
   C. Lower speed increases it
   D. Viscosity has no effect
   Answer: A. S = (mu N/P)(r/c)^2: higher viscosity and speed and lower load give a thicker film. (Key point: Sommerfeld number)

3. What is the coefficient of friction found from Petroff's equation here?
   A. 0.0247
   B. 0.247
   C. 0.00247
   D. 2.47
   Answer: A. P = 5x10^5 Pa, N = 25 rev/s, r/c = 1000, so f = 2 pi^2 (0.025x25/5x10^5)(1000) = 0.0247. (Key point: Petroff calculation of f)

4. How are friction torque and power lost obtained?
   A. T = f W r = 4.93 N m, power = 2 pi N T = about 775 W
   B. T = f W r = 4.93 N m, power = T/N = 0.2 W
   C. T = f W = 98.7 N m, power = 2 pi N T
   D. T = W r = 200 N m, power = f T
   Answer: A. T = 0.02467 x 4000 x 0.05 = 4.93 N m and power = 2 pi x 25 x 4.93, about 775 W. (Key point: Torque and power loss)

## Related in Internal Combustion Engines

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## Related in other subjects

- [Petroff equation for bearing friction](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Apetroff-equation-bearing-friction) — Machine Design I, chapter 7
- [Hydrodynamic lubrication theory and film formation](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Ahydrodynamic-lubrication-theory) — Machine Design I, chapter 7
- [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
- [Torque and power loss in a shaft bearing](https://www.bctengineer.com/question.md?key=semester_4%2Ffluid_mechanics_mechanical%3Ashaft-in-bearing-torque-power) — Fluid Mechanics, chapter 1