---
title: "Torque and power loss in a shaft bearing"
subject: "Fluid Mechanics"
chapter: 1
key: "semester_4/fluid_mechanics_mechanical:shaft-in-bearing-torque-power"
source: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/shaft-in-bearing-torque-power
site: BCT Engineering
---

# Torque and power loss in a shaft bearing

Fluid Mechanics · Chapter 1: Definition and Analysis method

- Asked in: year not given, 6 marks
- Video: [Problem to Determine the Power Lost in the Bearing (Sleeve)- Properties of Fluid - Fluid Mechanics 1](https://www.youtube.com/watch?v=2lhmSUnTckE) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/shaft-in-bearing-torque-power
- Whole chapter: https://www.bctengineer.com/md/semester_4/fluid_mechanics_mechanical/chapter-1

## Question

A shaft of diameter 100 mm rotates at 300 rpm inside a stationary sleeve (journal bearing) of length 150 mm. The uniform radial clearance between the shaft and sleeve is 0.2 mm and it is filled with oil of dynamic viscosity 0.08 Pa s. Assuming a linear velocity profile in the oil, find the shear stress on the shaft surface, the resisting torque and the power lost in viscous friction.

## Short answer

Using linear velocity profile, surface speed U=1.571 m/s, shear stress τ=μU/c=628 Pa. Resisting torque T=τ·πDL·D/2=1.48 N·m. Power lost P=Tω=46.5 W.

## Answer

#### Given data
$D = 0.1\ \text{m}$, $L = 0.15\ \text{m}$, $c = 0.2\ \text{mm} = 2\times10^{-4}\ \text{m}$, $N = 300\ \text{rpm}$, $\mu = 0.08\ \text{Pa s}$.

Since the clearance is very small compared with the diameter, the oil film can be treated as flat with a linear velocity profile, so $du/dy = U/c$.

#### Surface speed of the shaft
$$
U = \frac{\pi D N}{60} = \frac{\pi \times 0.1 \times 300}{60} = 1.571\ \text{m/s}
$$

#### Shear stress
$$
\tau = \mu \frac{U}{c} = 0.08 \times \frac{1.571}{2\times10^{-4}} = 628.3\ \text{Pa}
$$

#### Resisting torque
Shear area $A = \pi D L = \pi \times 0.1 \times 0.15 = 0.04712\ \text{m}^2$

$$
\begin{aligned}
F &= \tau A = 628.3 \times 0.04712 = 29.61\ \text{N} \\
T &= F \times \frac{D}{2} = 29.61 \times 0.05 = 1.480\ \text{N m}
\end{aligned}
$$

#### Power lost
$$
\begin{aligned}
\omega &= \frac{2\pi N}{60} = 31.42\ \text{rad/s} \\
P &= T\omega = 1.480 \times 31.42 = 46.5\ \text{W}
\end{aligned}
$$

**Answer: $\tau = 628\ \text{Pa}$, $T = 1.48\ \text{N m}$, $P = 46.5\ \text{W}$**

## Check yourself

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

1. Why can du/dy be taken as U/c in this journal bearing?
   A. Because the oil is a non-Newtonian fluid
   B. Because the clearance is very small compared with the diameter, so the film behaves like a flat layer with linear velocity profile
   C. Because the shaft rotates at constant rpm
   D. Because the sleeve also rotates at the same speed
   Answer: B. A very small clearance makes the film nearly flat, so velocity varies linearly from U at the shaft to zero at the sleeve. (Key point: Linear velocity profile in thin film)

2. The shaft surface speed is 1.571 m/s and the clearance is 0.2 mm. The shear stress on the shaft is found from:
   A. tau = mu x c / U = 0.08 x 0.0002/1.571
   B. tau = mu x U / c = 0.08 x 1.571/0.0002 = 628 Pa
   C. tau = U / (mu c)
   D. tau = mu x U x c
   Answer: B. tau = mu U/c = 0.08 x 1.571 / 2e-4 = 628.3 Pa. (Key point: Shear stress calculation)

3. After finding tau = 628.3 Pa, how are the torque and power lost obtained?
   A. F = tau x (pi D L), T = F x D/2, P = T x omega
   B. F = tau x (pi D^2/4), T = F x D, P = T x N
   C. T = tau x D, P = T / omega
   D. F = tau / (pi D L), T = F x D/2, P = F x omega
   Answer: A. Shear force is stress times the cylindrical surface area pi D L (0.04712 m^2), torque is force times radius (1.48 N m), and power is T x omega = 46.5 W. (Key point: Torque and power)

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