# What is the difference between performance and efficiency

## Work, energy and performance

#### Maximum efficiency

From the **Conservation of energy law** it follows that the amount of useful energy can never be greater than the amount of energy supplied. This means that the efficiency of an energy converter cannot be greater than 1. One also says: The efficiency cannot be greater than 100%.

In practice, losses occur in every real energy converter, e.g. due to friction. \ (\ Delta E _ {\ rm {util}} \) is therefore smaller than \ (\ Delta E _ {\ rm {zu}} \). Thus the efficiency is \ (\ eta <1 \) or less than 100%. Therefore, a perpetual motion machine cannot be built in practice.

You can find information about typical levels of efficiency, e.g. of motors and generators, in the outlook.

#### Definition of performance

Since the relationship \ (\ Delta E = P \ cdot \ Delta t \) applies to energy and power, you can also write the efficiency \ (\ eta \) as the ratio of the corresponding powers: \ [\ eta = \ frac {{\ Delta {E _ {\ rm {us}}}}} {{\ Delta {E _ {\ rm {zu}}}}} = \ frac {{{P _ {\ rm {us}}} \ cdot \ Delta t}} {{{P _ {\ rm {to}}} \ cdot \ Delta t}} \ Rightarrow \ eta = \ frac {{{P _ {\ rm {useful}}}}} {{{P _ {\ rm {to}}}}} \]

#### Overall efficiency with several conversions in a row

If several energy conversions take place one after the other in an energy converter, you can calculate the overall efficiency by multiplying the efficiency of the individual conversions together.

In the electric motor of a drilling machine, 30% of the electrical energy supplied is lost (heat and friction losses). In the transmission, 10% of the mechanical energy supplied by the motor is lost due to friction. Thus the efficiency of the engine \ (\ eta _ {\ rm {engine}} = 70 \% = 0 {,} 7 \) and that of the gearbox \ (\ eta _ {\ rm {gear}} = 90 \% = 0 {,} 9 \).

The following then applies to the overall efficiency of the hammer drill: \ [\ eta _ {\ rm {Drill}} = \ eta _ {\ rm {Motor}} \ cdot \ eta _ {\ rm {Gear}} \] \ [\ Rightarrow \ eta _ {\ rm {drill}} = 0 {,} 70 \ cdot 0 {,} 90 = 0 {,} 63 = 63 \% \] 63% of the supplied electrical energy is converted into mechanical rotary energy of the drill head.

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