4.03 Rates in the home

Comparing rates, as we first saw in 4.02, is also useful when considering the costs of purchasing and owning an electrical appliance. There are many brands and types on offer, so it can be difficult to figure out which purchase is best for your budget and personal needs. One thing to consider is the rate of energy consumption of each appliance, which will cost money in the long run, in addition to the upfront costs.

The rate of energy consumption or power of an appliance is often given in watts (W) and is a standard unit of measurement that describes the amount of energy (in Joules) consumed per second. In the case that the power of the appliance is really high, it may be more convenient to report the power in kilowatts (kW) where $1$1 kW represents $1000$1000 W.

One more important unit of measure is the kilowatt hour, written as kWh. One kilowatt hour is the amount of energy an appliance consumes at a rate of one kilowatt for one hour. Mathematically, this means that one kilowatt hour is the rate in kilowatts multiplied by the number of hours. Working with energy units lets us identify the energy efficiencies and costs of electrical appliances.

Exploration

The cost and power of two fridges from two competing brands are given in the table below. Whilst the price of Electricool is less than IcePower, the Electricool fridge uses more power to run. Intuition tells us that the cost to run the Electricool fridge will be more than the cost to run the IcePower fridge, since more power is used. This can be calculated if we know the cost of electricity.

The average rate for domestic electricity is $\$0.15$$0.15/kWh. So to find the annual cost to run an Electricool fridge, we want to find the amount of energy (in kWh) it uses per year. We first want to convert the power of the Electricool fridge from watts to kilowatts by dividing by $1000$1000. So the Electricool fridge operates at $2.5$2.5 kW. The amount of energy consumed by the fridge in one hour is $2.5$2.5 kWh. This is because we multiply the power (in kW) by time (in hours). Since there are $24$24 hours in a day and $365$365 days in a year, the total amount of energy used to operate the Electricool fridge is $2.5\times24\times365=21900$2.5×24×365=21900 kWh. Each one of these kilowatt hours costs $\$0.15$$0.15, so overall it costs $21900\times\$0.15=\$3285$21900×$0.15=$3285.

Finding the annual cost to run an IcePower fridge follows the same procedure. The fridge operates at $2$2 kW and consumes $2\times24\times365=17520$2×24×365=17520 kWh in a year. The overall running cost is $17520\times\$0.15=\$2628$17520×$0.15=$2628.

So in a year, the difference in running costs is $\$3285-\$2628=\$657$$3285−$2628=$657. One important thing to notice is that the difference in running costs after one year is more than the difference in the purchase price of $\$400$$400. So if we purchased the initially more expensive IcePower fridge, we know that after a year we actually save money.

In fact, if we divide the purchase difference of $\$400$$400 by the savings we make in running costs, we get the following value: $\frac{400}{657}\approx0.61$400657​≈0.61. This tells us how long it takes in years before the savings we make in running costs make up for the extra costs of purchasing the expensive fridge - this works out to just over seven months.

Practice questions

question 1

question 2

question 3