estimating_cost_of_electricity

Estimating Cost of Electricity

It turns out that in BC, there's an incredibly easy way to estimate the cost of electricity for devices that are always on 24/7:

Take the power consumption in Watts (W)
Divide by 10
You get the cost of electricity per month for that device

That's right, if something draws 50 W and it's on 24/7, it costs about $5 per month of electricity to run it.

Here's why that works...

In BC, the price of electricity is 10.97 ¢/kWh for the first tier and 14.08 ¢/kWh for the second tier. So let's take the worst case scenario and imagine that we're always on tier 2 pricing. The math is just a string of multiplication to cancel out units:

\begin{align*} &\frac{365\text{ days}}{\text{year}} \times \frac{24\text{ hrs}}{\text{day}} \times \frac{1\text{ year}}{12\text{ months}} \times 14.08\frac{\text{ ¢}}{\text{kWh}} \times \frac{1 \text{ \$}}{100\text{ ¢}} \times \frac{1\text{ kW}}{1000\text{ W}} \\
=\quad& 0.102784 \ \tfrac{\text{\$}}{\text{month}\cdot\text{W}}
\end{align*} If you look at the units carefully, you'll notice that they almost all cancel out except for \$\frac{\text{\$}}{\text{month}\cdot\text{W}}\$ ¹⁾

That number means that it costs $0.102784 per month for each Watt of power that's continuously drawn. So if you multiply that number by the power consumption of the device, the \$\text{W}\$ will cancel out and you'll get a result in \$\frac{\text{\$}}{\text{month}}\$, which is the monthly electricity price for that device. For example:

\$$ 50 \text{ W} \times 0.102784 \ \tfrac{\text{\$}}{\text{month}\cdot\text{W}} = 5.1392 \ \tfrac{\text{\$}}{\text{month}} \$$

The trick to simply divide by 10 instead of multiplying by 0.102784 works because \$\tfrac{1}{0.102784} = 9.729... \approx 10\$ . So a better approximation would be to divide by 9.7, but dividing by 10 is so much easier to do in your head, and the result is not that far off, specially since we already over estimated the cost of electricity to full tier 2 instead of a combination of tier 1 and tier 2.

A Concrete Example

At home, I run a bunch of equipment²⁾ on a dedicated 12V system. The system is fed with a high quality, high power 12 V charger with backup batteries and solar panels. In the winter, the solar panels don't get any sun and the charger puts out a constant 75 W (which goes up when I'm transmitting). In the summer, the solar panels provide all the electricity needed for up to 16 hours.

That means that my internet and radio system costs me about $7.50 / month of electricity in the winter, a third of that in the summer.

¹⁾

The “hrs” of “24 hrs” cancels with the “h” of “kWh” since kWh means kW \$ \times\$ hr .

²⁾

Equipment on dedicated 12 V system that's on 24/7:

Internet modem, switches, and Wifi Boosts
Echolink radio and computer
AREDN hAP, two AREDN dishes, one AREDN VOIP Phone

Sun Coast Amateur Radio Club Society - blog:2024-09-28

estimating_cost_of_electricity

Estimating Cost of Electricity

Here's why that works...

A Concrete Example

Sun Coast Amateur
Radio Club Society - blog:2024-09-28