Energy efficiency

​​In a nutshell

The efficiency of an energy transfer is how much useful energy is transferred by the system compared to the total energy supplied to the system. The efficiency can be improved by reducing friction, thermal conductivity, electrical resistance and air resistance.

Equations

Variable definitions

Efficiency

Definition

The ratio of useful energy transferred by the system to the total energy supplied to the system.

During energy transfers, some energy is wasted or dissipated and the rest is transferred for useful purposes. Energy is wasted when it is converted to energy stores that aren't the intended use.

Example

A lamp's purpose is to provide light to its surroundings. However, the bulb also heats up as some of the electrostatic energy is converted to thermal energy. This energy is wasted as it is not useful.

A system with a high efficiency means most of the energy transferred is useful. A system with a low efficiency means most of the energy transferred is wasted.

Efficiency can be calculated using the equation:

​$$efficiency = \dfrac{useful\space output \space energy \space transferred}{total\space input\space energy\space transferred}$$​​

This gives efficiency as a value between $$0$$ and $$1$$.

Note: The efficiency can be converted into a percentage by multiplying the equation above by $$100\,\%$$.

Alternatively, the equation for efficiency can be written in terms of power (as $$P = \frac{E}{t}$$):

​$$efficiency = \dfrac{useful\space power \space output }{total\space power\space input}$$​​

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Example

A microwave has an efficiency of $$50\,\%$$. It has a total power of $$800\,W$$. How much energy is wasted in $$3 \,minutes$$?

First, write out the quantities given in the question and check they're in the correct form:

$$efficiency = 50\,\% = \frac{50\,\%}{100\,\%} = 0.5\\ \ \\ P = 800\,W\\ \ \\ t = 3\,mins = 3 \times 60 = 180\,s$$​​

Next, write down the equations needed and rearrange:

$$efficiency = \dfrac{useful\space power \space output }{total\space power\space input}\\ \ \\useful\space power \space output = efficiency \times total\space power\space input\\ \ \\ wasted \space power \space output= total \space power \space input - useful \space power \space output\\ \ \\P = \dfrac{E}{t} \to E = P\times t$$​​

Then, substitute the values into the equations:

$$useful \space power \space output = 0.5 \times 800 = 400\,W\\ \ \\ wasted \space power \space output = 800 - 400 = 400\,W\\ \ \\ E = 400 \times 180 = 72\,000 = 7.2 \times 10^4$$​​

Make sure to include the correct units:

$$energy\ wasted, E = 7.2 \times 10^4\,J$$​​

The energy wasted is $$\underline{7.2\times10^4 \,J}$$.​

Increasing efficiency

To improve the efficiency of a system, the energy that is wasted or dissipated needs to be reduced. Methods to do this vary based on the type of energy that's wasted.

Tip: Poor thermal conductors are good thermal insulators! Metals are good thermal conductors, which is why houses aren't made out of metals.

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Example

Hot liquids can be kept in flasks to reduce the rate at which they cool down. This is done by having vacuum surrounding the container. 

1. Plastic stopper 2. Glass walls with silver coating 3. Vacuum 4. Plastic spacer

The thermal conductivity of a vacuum is zero so heat cannot be transferred out of the system through the vacuum. Instead a reduced amount of heat is transferred out through the plastic stopper and spacer. Therefore, the rate at which the liquid cools down is significantly decreased than a normal bottle!