Power and efficiency

In a nutshell

Power is the rate at which energy is transferred or work is done. Efficiency is the measure of how much useful energy a machine outputs compared to the total energy supplied.

Equations

DESCRIPTION	EQUATION
Power equation	$P = \dfrac{W}{t}=Fv$
Efficiency equation	$Efficiency \ (\%) = \dfrac{Useful \ Output \ Energy}{Total \ Input \ Energy} \times 100$

Variable definitions

QUANTITY NAME	SYMBOL	DERIVED UNIT	SI UNIT
$power$	$P$	$W$	$kg m^2s^{-3}$
$force$	$F$	$N$	$kgms^{-2}$
$time$	$t$	$s$	$s$
$Work \ done$	$W$	$J$	$m^2kgs^{-2}$
$velocity$	$v$	$ms^{-1}$	$ms^{-1}$

Power

Power is the rate at which energy is transferred. It can be expressed as:

$P = \dfrac{W}{t}$

As work done is equal to energy transfer, power can also be expressed as the rate of work done.

Power is measured in watts. 1 watt is equal to one joule per second.

Example:

A man of mass $70 \ kg$ runs for $20 \ minutes$ at a speed of $5 \ ms^{-1}$ . Calculate the rate of work done on the man.

Firstly state the variables from the question:

$m = 70 \ kg, \ \ \ \ t = 20 \times 60 = 1200s, \ \ \ \ v= 5 \ ms^{-1}$

Next state the equation needed:

$P= \dfrac{W}{t}, \ \ E_k = \dfrac{1}{2}mv^2, \ \ E_k = W$

Rearrange and derive new equation for power:

$P = \dfrac{\frac{1}{2}mv^2}{t}$

Next substitute in the values and solve:

$P = \dfrac{\frac{1}{2}70 \times (5)^2}{1200}\newline \\[0.1in] P = 0.729 \ W\newline \\[0.1in] P = 0.73 \ W (2sf)$

The rate of work done of the man is $\underline{0.73 \ W}.$

Power to move at a constant velocity

The equation for power can be used to derive an equation for the power required to move an object at constant velocity against resistive forces. In order for an object to move at a constant velocity, the net forces and net acceleration must be zero. This means the driving force of the object must be equal to the resistive forces acting on an object.

The driving force of a car, for example, can be used when considering the power required for the car to maintain a constant velocity. This can be expressed as:

$P= \dfrac{W}{t}$

Work done can be expressed as:

$W=Fx$

You can then substitute in the equation for velocity:

$v=\dfrac{x}{t}$

Substituting into the equation for power:

$P = \dfrac{Fx}{t}$

As you know $v = \frac{x}{t},$ you can obtain:

$P= Fv$

Efficiency

The conservation of energy states that energy is constant in a closed loop. There are 'energy losses' in the real world, such as electrical devices producing thermal energy. The total energy, however is always conserved.

Efficiency is the measure of how much useful energy a machine outputs compared to the total energy supplied. It can be expressed mathematically as:

$Efficiency \ (\%) = \dfrac{Useful \ Output \ Energy}{Total \ Input \ Energy} \times 100$

Note: efficiency can be expressed as either a percentage or a decimal. It can never be above $100 \ \%$ or $1$ , as this violates the conservation of energy.

Example

When a ball is dropped (with no added force except that of its weight), the ball will bounce, but will never return to it's original height. This is because the gravitational potential energy of the ball is transferred into wasted energy, such as thermal energy, as well as useful energy. You could work out the efficiency of the ball by dividing the bounce height by the dropped height.

Physics; Energy; KS5 Year 12; Power and efficiency

Example:

An LED has an output energy of $0.1 \ MJ$ . It dissipates $0.065 \ MJ$ as thermal energy and converts the rest into light energy. What is the efficiency of the LED?

First state the variables from the question:

$Useful \ energy \ output = 0.1-0.065 = 0.035 \ J\newline Total \ energy \ output = 0.1 \ J$

Then state the equation needed:

$Efficiency \ (\%) = \dfrac{Useful \ Output \ Energy}{Total \ Input \ Energy} \times 100$

Substitute into the equation and solve:

$Efficiency \ (\%) = \dfrac{0.035}{0.1} \times 100\newline \\[0.1in]Efficiency \ (\%) = 35 \%$

The efficiency of the LED is $\underline{35\%}$ .