Sun Coast Amateur
Radio Club Society

User Tools

Site Tools

You are here: home » blog » 2024-09-28 » estimating_cost_of_electricity
blog:2024-09-28:estimating_cost_of_electricity

Differences

This shows you the differences between two versions of the page.

Link to this comparison view

Both sides previous revisionPrevious revision
Next revision		Previous revision
blog:2024-09-28:estimating_cost_of_electricity [2024/09/28 10:04] – va7fiblog:2024-09-28:estimating_cost_of_electricity [2024/11/24 12:48] (current) – va7fi
Line 1: Line 1:
 +~~DISCUSSION~~
 +~~NOTOC~~
 +====== 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:
 +<WRAP center round box 60%>
 +  * Take the power consumption in Watts (W)
 +  * Divide by 10
 +  * You get the cost of electricity per month for that device
 +</WRAP>
 +
 +That's right, if something draws 50 W and it's on 24/7, it costs about $5 per month of electricity to run it.
 +
 +===== Here's why that works... =====
 +
 +In BC, the [[https://app.bchydro.com/accounts-billing/rates-energy-use/electricity-rates/residential-rates/tiered.html |price of electricity]] is 10.97 ¢/kWh for the first tier and 14.08 ¢/kWh for the second tier.  So let's take the worst case scenario and imagine that we're always on //tier 2// pricing.  The math is just a string of multiplication to cancel out units:
 +
 +\begin{align*}
 +&\frac{365\text{ days}}{\text{year}}
 +\times \frac{24\text{ hrs}}{\text{day}}
 +\times \frac{1\text{ year}}{12\text{ months}}
 +\times 14.08\frac{\text{ ¢}}{\text{kWh}}
 +\times \frac{1 \text{ \$}}{100\text{ ¢}}
 +\times \frac{1\text{ kW}}{1000\text{ W}} \\\\
 +=\quad& 0.102784 \ \tfrac{\text{\$}}{\text{month}\cdot\text{W}}\\
 +\end{align*}
 +If you look at the units carefully, you'll notice that they almost all cancel out except for \$\frac{\text{\$}}{\text{month}\cdot\text{W}}\$ ((The "hrs" of "24 hrs" cancels with the "h" of "kWh" since kWh means kW \$ \times\$ hr .))
 +
 +That number means that it costs %%$%%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 equipment((
 +Equipment on dedicated 12 V system that's on 24/7:
 +  * Internet modem, switches, and Wifi Boosts
 +  * [[https://scarcs.ca/links/echolink#sysop |Echolink radio and computer]]
 +  * [[https://wcaredn.ca/setups/va7fi/home |AREDN hAP, two AREDN dishes, one AREDN VOIP Phone]]
 +))
 +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.