What is efficiency in physics

Efficiency has multiple definitions, all of them valid. We can define efficiency as:

  • the ability to avoid wasting energy when performing a particular work
  • the ratio between the useful work performed by a device and the total energy consumed as input

Let’s suppose we have a system which receives a power as input and outputs another power. The efficiency is the ratio between the output and input power.

Efficiency of a system

Image: Efficiency of a system

The symbol used to define efficiency is the Greek letter eta (η):

\[ \begin{equation} \begin{split}
\end{split} \end{equation} \]

If we want to express the efficiency as percentage the mathematical expression becomes:

\[\eta = \frac{P_{out}}{P_{in}} \cdot 100 [\%] \]

For example if we take an electric motor which receives a 1000 W power from a battery and outputs 900 W at the rotor, what is the efficiency of the motor?

\[\eta_{mot} = \frac{900}{1000} \cdot 100 = 90 \% \]

Where are the remaining 100 W gone? Why aren’t they available at the motor output (rotor)?

The answer is simple. Since the rotor is mounted on some bearings, there is an amount of friction in the bearings. The friction absorbs a part of the input power and transforms it into heat. Also there are some winding losses in the motor itself. The friction losses together with the winding losses reduces the output power of the motor.

\[P_{out} = P_{in} – P_{loss}\]

If we divide the above expression to input power we get:

\[ \begin{equation*} \begin{split}
\frac{P_{out}}{P_{in}} &= \frac{P_{in}}{P_{in}} – \frac{P_{loss}}{P_{in}}\\
\eta &= 1 – \frac{P_{loss}}{P_{in}}\\
\end{split} \end{equation*} \]

If we know the input power of a system and its efficiency, we can easily calculate the output power as:

\[ \begin{equation} \begin{split}
\bbox[#FFFF9D]{P_{out} = \eta \cdot P_{in}}
\end{split} \end{equation} \]

Now we are going to work on a example which will highlight the impact of efficiency on the output of an actuation system. Also we will see how to use efficiency for the calculation of output power.

Suppose we have an electro-mechanical actuation system composed by:

  • a battery
  • an electric motor
  • a worm gear
  • a spur gear
Electro-mechanical actuation system

Image: Electro-mechanical actuation system

By knowing the voltage and electrical current of the battery and the efficiency of motor, worm gear and spur gear, we can calculate the output power at the spur gear.

Physical variable Symbol Value Unit
Battery voltage \[U_{bat}\] 12 V
Battery current \[I_{bat}\] 10 A
Motor efficiency \[\eta_{mot}\] 95 %
Worm gear efficiency \[\eta_{worm}\] 70 %
Spur gear efficiency \[\eta_{spur}\] 98 %

For a better understanding of the input and output power for each component, we can describe the above actuation system with block diagrams:

Electro-mechanical actuation system block diagram

Image: Electro-mechanical actuation system block diagram

First we calculate the input power, the battery power:

\[P_{bat} = U_{bat} \cdot I_{bat} = 12 \cdot 10 = 120 W \]

Next we calculate the motor output power:

\[P_{mot} = \eta_{mot} \cdot P_{bat} = 0.95 \cdot 120 = 114 W \]

Next we calculate the worm gear output power:

\[P_{worm}=\eta_{worm} \cdot P_{mot} = 0.70 \cdot 114 = 79.8 W \]

Finally we calculate the spur gear output power:

\[P_{spur} = \eta_{spur} \cdot P_{worm} = 0.98 \cdot 79.8 = 78.204 W\]

By knowing the input power and the output power we can calculate the overall efficiency of the system:

\[\eta = \frac{P_{out}}{P_{in}} = \frac{P_{spur}}{P_{bat}} = \frac{78.204}{120}=0.6517 = 65.17 \% \]

The overall system efficiency can be also calculated by multiplying together all the efficiency of the components:

\[\eta = \eta_{mot} \cdot \eta_{worm} \cdot \eta_{spur} = 0.95 \cdot 0.70 \cdot 0.98 = 0.6517 = 65.17 \% \]

We can calculate the power losses by subtracting the output power from the input power:

\[P_{loss} = P_{in} – P_{out} = P_{bat} – P_{spur} = 120 – 78.204 = 41.796 W \]

With the power losses we can recalculate the overall efficiency as:

\[\eta = 1 – \frac{P_{loss}}{P_{in}} = 1 – \frac{41.796}{120} = 1 – 0.3483 = 0.6517 = 65.17 \%\]

With this exercise should be pretty obvious how the efficiency is calculated and it’s impact on the power output of a system.

For any questions or observations regarding this tutorial please use the comment form below.

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