The amount of usable horsepower at the crankshaft is called

Check out the new look and enjoy easier access to your favorite features

Engine power is the power that an engine can put out. It can be expressed in power units, most commonly kilowatt, pferdestärke (metric horsepower), or horsepower. In terms of internal combustion engines, the engine power usually describes the rated power, which is a power output that the engine can maintain over a long period of time according to a certain testing method, for example ISO 1585. In general though, an internal combustion engine has a power take-off shaft (the crankshaft), therefore, the rule for shaft power applies to internal combustion engines: Engine power is the product of the engine torque and the crankshaft's angular velocity.

Engine power

Common symbols

P

SI unitKilowatt (kW)In SI base units1000 kg⋅m2⋅s−3

Derivations from
other quantities

P = M\cdotω

Dimension

M L 2 T − 3 {\displaystyle ML^{2}T^{-3}}

Power is the product of torque and angular velocity:[1]

Let:

$P = {\displaystyle P=}$ Power in Watt (W)
$M = {\displaystyle M=}$ Torque in Newton-metre (N·m)
$n = {\displaystyle n=}$ Crankshaft speed per Second (s−1)
$ω = {\displaystyle \omega =}$ Angular velocity = $2 π n {\displaystyle 2\pi n}$

Power is then:

P = M ⋅ ω {\displaystyle P=M\cdot \omega }

In internal combustion engines, the crankshaft speed $n {\displaystyle n}$ is a more common figure than $ω {\displaystyle \omega }$ , so we can use $2 π n {\displaystyle 2\pi n}$ instead, which is equivalent to $ω {\displaystyle \omega }$ :[2]

P = M ⋅ 2 π ⋅ n {\displaystyle P=M\cdot 2\pi \cdot n}

Note that $n {\displaystyle n}$ is per Second (s−1). If we want to use the common per Minute (min−1) instead, we have to divide $n {\displaystyle n}$ by 60:

P = M ⋅ 2 π ⋅ n 60 {\displaystyle P=M\cdot 2\pi \cdot {n \over 60}}

The approximate numerical value equations for engine power from torque and crankshaft speed are:[1][3][4]

International unit system (SI)

Let:

$P = {\displaystyle P=}$ Power in Kilowatt (kW)
$M = {\displaystyle M=}$ Torque in Newton-metre (N·m)
$n = {\displaystyle n=}$ Crankshaft speed per Minute (min−1)

Then:

P = M ⋅ n 9550 {\displaystyle P={M\cdot n \over 9550}}

Technical unit system (MKS)

$P = {\displaystyle P=}$ Power in Pferdestärke (PS)
$M = {\displaystyle M=}$ Torque in Kilopondmetre (kp·m)
$n = {\displaystyle n=}$ Crankshaft speed per Minute (min−1)

Then:

P = M ⋅ n 716 {\displaystyle P={M\cdot n \over 716}}

Imperial/U.S. Customary unit system

$P = {\displaystyle P=}$ Power in Horsepower (hp)
$M = {\displaystyle M=}$ Torque in Pound-force foot (lbf·ft)
$n = {\displaystyle n=}$ Crankshaft speed in Revolutions per Minute (rpm)

Then:

P = M ⋅ n 5252 {\displaystyle P={M\cdot n \over 5252}}

Example

Torque and power diagram of the example diesel engine

The power curve (orange) can be derived from the torque curve (blue)
by multiplying with the crankshaft speed and dividing by 9550

A diesel engine produces a torque $M {\displaystyle M}$ of 234 N·m at $n {\displaystyle n}$ 4200 min−1, which is the engine's rated speed.

Let:

$M = 234 N ⋅ m {\displaystyle M=234\,N\cdot m}$
$n = 4200 m i n − 1 = 70 s − 1 {\displaystyle n=4200\,{min}^{-1}=70\,s^{-1}}$

Then:

234 N ⋅ m ⋅ 2 π ⋅ 70 s − 1 = 102 , 919 N ⋅ m ⋅ s − 1 ≈ 103 k W {\displaystyle 234\,N\cdot m\cdot 2\pi \cdot 70\,s^{-1}=102,919\,N\cdot m\cdot s^{-1}\approx 103\,kW}

or using the numerical value equation:

234 ⋅ 4200 9550 = 102.91 ≈ 103 {\displaystyle {234\cdot 4200 \over 9550}=102.91\approx 103}

The engine's rated power output is 103 kW.

	Kilowatt	Kilopondmetre per Second	Pferdestärke	Horsepower	Pound-force foot per minute
1 kW (= 1000 kg·m2·s−3) =	1	101.97	1.36	1.34	44,118
1 kp·m·s−1 =	0.00980665	1	0.013	0.0132	433.981
1 PS =	0.73549875	75	1	0.986	32,548.56
1 hp =	0.7457	76.04	1.014	1	33,000
1 lbf·ft·min−1 =	2.26·10−5	0.0023	2.99·10−5	3.03·10−5	1