Engine power

 ( Mechanics ) 

Engine power is the power that an engine can develop. It can be expressed in power units, most commonly kilowatt, metric horsepower (often abbreviated PS), 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.

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Definition Power is the product of torque and angular velocity:. p 233

Let:

• $P=$ Power in Watt (W)
• $M=$ Torque in Newton-metre (N·m)
• $n=$ Crankshaft speed per Second (s⁻¹)
• $\backslash omega=$ Angular velocity = $2\backslash pi\; n$

Power is then:

$P=\; M\; \backslash cdot\; \backslash omega$

In internal combustion engines, the crankshaft speed $n$ is a more common figure than $\backslash omega$, so we can use $2\; \backslash pi\; n$ instead, which is equivalent to $\backslash omega$:. p. 21

$P=\; M\; \backslash cdot\; 2\; \backslash pi\; \backslash cdot\; n$

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

$P=\; M\; \backslash cdot\; 2\; \backslash pi\; \backslash cdot\; \{n\; \backslash over\; 60\}$

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Usage

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Numerical value equations The approximate numerical value equations for engine power from torque and crankshaft speed are:. p 154. p 259

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International unit system (SI)

Let:

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

Then:

$P=\; \{M\; \backslash cdot\; n\; \backslash over\; 9550\}$

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Technical unit system (MKS)

• $P=$ Power in Metric horsepower (hp, PS)
• $M=$ Torque in Kilopondmetre (kp·m)
• $n=$ Crankshaft speed per Minute (min⁻¹)

Then:

$P=\; \{M\; \backslash cdot\; n\; \backslash over\; 716\}$

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Imperial/U.S. Customary unit system

• $P=$ Power in Horsepower (hp, bhp)
• $M=$ Torque in Pound-force foot (lbf·ft)
• $n=$ Crankshaft speed in Revolutions per Minute (rpm)

Then:

$P=\; \{M\; \backslash cdot\; n\; \backslash over\; 5252\}$

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Example

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

Let:

• $M=\; 234\; \backslash ,\; N\; \backslash cdot\; m$
• $n=\; 4200\; \backslash ,\; \{min\}^\{-1\}\; =\; 70\; \backslash ,\; s^\{-1\}$

Then:

$234\; \backslash ,\; N\; \backslash cdot\; m\; \backslash cdot\; 2\; \backslash pi\; \backslash cdot\; 70\; \backslash ,\; s^\{-1\}\; =\; 102,919\; \backslash ,\; N\; \backslash cdot\; m\; \backslash cdot\; s^\{-1\}\; \backslash approx\; 103\; \backslash ,\; kW$

or using the numerical value equation:

$\{234\; \backslash cdot\; 4200\; \backslash over\; 9550\}\; =\; 102.91\; \backslash approx\; 103$

The engine's rated power output is 103 kW.

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Units

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See also

• List of production cars by power output

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Bibliography

Categories: Mechanics, Power, Engines

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