Snow MeltingFrequently Asked Questions
Hopefully, by browsing this page and clicking on the questions below, you should be able to find an answer to your Snow Melting related problem. If that is not the case then you can visit our forum or let us help you, either by using our live Chat or Ask A Question services.
To calculate the system's approximate operating costs, multiply the total kilowatts of the system by the cost per kilowatt in your area. Let's use the example of a 350 sq. ft. of concrete patio (residential application), with 240 VAC.
To determine Total Watts: multiply area in sq. ft. x 50 Watts
350 x 50 = 17,500 Total Watts
To arrive at Kilowatts (the unit in we purchase electricity): take the Total Watts and divide by 1000
17,500 Total Watts / 1000 = 17.5 Kilowatts.
To determine Kilowatt rate: use the national average of .12 cents per Kilowatt-hours so how much do we pay for 17.5 kWatts?
17.5 kW x .12 = $2.10 for every full hour of operation.
How many hours would you operate the system? We use a typical 6 hour snowfall for our example so
$2.10 x 6 hours = $12.60 for that snowfall.
All of this will vary due to "after-run time". "After run time" is where the system remains on extra hours after the snow has completed falling, to ensure complete pavement snow melting and drying.
Power requirements are measured in Amps and based on three factors: the watts per sq. ft. of the snow melting cable or mat, the area powered and the voltage used for the application. Our product is rated at 50 watts per sq. ft., so that is a constant factor. Most snowmelt projects are powered with 240V AC – not all, but most. The one factor that is always variable is the area to be heated/powered. That is based on your project. For a point of reference, let’s use the example above of a 350 sq. ft. concrete patio.
The formula is (50 W/sq. ft. x area of 350 sq. ft.) / V (240V) = Amps, so 17,500 / 240V = 72.9 Amps