Engineer's Notebook §1.1

Power from torque

A rotating shaft delivers power equal to its torque times its angular speed. Engine power figures come straight from this:

P = Tω
P
power
W
T
torque
N·m
ω
angular speed
rad/s
N
rotational speed
rpm
G
gear ratio
—
η
efficiency
—
F
tractive force
N
r
wheel radius
m
Power against engine speed for a constant torque: a straight line from the origin. A point marks the chosen engine speed. ENGINE SPEED N, rpmPOWER P, kW 9,000 P = 126 kW 500
Fig. 1.1-1At constant torque, power rises in proportion to speed. Real engines follow a torque curve.
ω = 2πN/60419 rad/s
P in horsepower169 hp
P = Tω126 kW

Worked example

Uses the values in Fig. 1.1-1. Change them and the working updates.

Given
  • Tshaft torque300 N·m
  • Nengine speed4,000 rpm
Find

The power delivered P.

Solution
ω = 2πN/60 = 2π × 4,000 rpm / 60 = 419 rad/s P = Tω = 300 N·m × 419 rad/s = 126 kW
Answer
P = 126 kW

That is 169 hp (1 hp = 745.7 W).

Related forms

  • Through a gearbox, ratio G

    Tout = GηTinωout = ωinG

  • Tractive force at the wheel

    F = Twheelr

  • In everyday units

    P [kW] = T [N·m] × N [rpm]9549

Assumptions
  • The shaft turns at a steady speed.
  • T is the torque measured at the shaft (brake torque).
  • The demo assumes the same torque at every speed to show the proportionality. Real engines follow a torque curve.
  • No losses. Power through a gearbox is reduced by its efficiency.

Derivation
  1. A torque turning a shaft through angle θ does work W = Tθ.
  2. Power is the rate of doing work: P = dW/dt = T dθ/dt = Tω.
  3. Engine speed in rpm converts to angular speed as ω = 2πN/60.

1 mechanical horsepower is 745.7 W. The 9549 in the everyday form is 60,000/2π.