GBA Logo horizontal Facebook LinkedIn Email Pinterest Twitter Instagram YouTube Icon Navigation Search Icon Main Search Icon Video Play Icon Plus Icon Minus Icon Picture icon Hamburger Icon Close Icon Sorted

Community and Q&A

PV panels required to power electrical boiler to replace propane boiler

minpix | Posted in Energy Efficiency and Durability on

I am considering replacing my 25 year old propane boiler (which I’m told is at most 80% efficient)  used to heat my hydronic floors, with a new, high efficiency, electric boiler to heat both the floors and DHW, and powering it with photovoltaic panels. Before I get too deep into it though, I want to guesstimate how many PV panels would be required to run this system.

My current propane consumption is about 1400 gallons per year. We use propane for 1) hydronic floor heating 2) 50 gal DHW tank 3) stove top cooking 4) clothes dryer and 5) hot tub. The hot tub is only cranked up perhaps a dozen times a year and the clothes drier is used 1-2 a week. Online estimates suggest 90-95% of my total consumption would have been used on DHW & floor heat. 

My question is therefore, how big a PV solar array would be required to deliver the same amount of heat from a +/- 95% efficiency electric boiler as is currently produced by burning 1300 gallons of propane in an 80% efficient gas boiler.

Thanks much!

GBA Prime

Join the leading community of building science experts

Become a GBA Prime member and get instant access to the latest developments in green building, research, and reports from the field.


  1. walta100 | | #1

    1400 x .9 = 1260 gallons of propane used for heating. 1260 x 91,330 = 24355800 BTU burned. 24355800 x.8 = 19484640 BTUs extracted. 19484640 \ 3412 = 5710.6 kWh

    5710 kWh in terms of # of solar panels depends on your location on earth and your local weather.

    All resistance boilers should be 100% efficient.

    You may want to consider a heat pump it would likely use 33% of the kWh that the boiler would require.


    1. Jon_R | | #3

      > 1260 x 91,330 = 24355800 BTU

      You should review your math.

  2. Expert Member
    Akos | | #2

    Extending on Walta's calc, around me a 1kW array produces 1100kWh/year. So based on that resitance only you would need a relatively modest ~5.2kW array.

    The problem is the array would produce a lot of power in the summer an very little in the winter when you would need most of the energy. If you are on a yearly net metered contract this doesn't matter.

    If yearly net metering is not an option, you would need a fair bit larger array to meet your winter heating needs. This usually ends up being too large for most houses. At that point it is usually best to look at reducing your load (more insulation, high efficiency heat pump heating).

    If you are installing the array it is also good to account on your house's plug load usage as this is usually a fair bit of energy as well.

  3. Expert Member
    BILL WICHERS | | #4

    Jon R points out that Walta’s eatimate is off by a factor of about 4.7x, so really need about 26,837 kWh, assuming the rest of his calcs are accurate. That’s a big difference.

    You also need to keep in mind that your heat loss will be highest at night when the solar output will be zero. If, for example, 2/3 of your heat load is at night, you need 3x the expected amount of solar output to cover the entire day, and you need to store 2/3 of the output for use at night. None of the conversions to battery and back is 100% efficient, so you actually need even more solar output than you’d think from my simple example. Note that this is just a conceptual example, not based on any hard numbers. Battery storage is EXPENSIVE and also a maintenance item. If you have a grid tied system then energy storage isn’t an issue, but you still need to oversize your solar system if you want to come out to net zero on your utilities.

    To give much more info, we’d need to know your climate zone and about where in the country you live. Somewhere with cloudy winters like the Midwest is not nearly as good for solar potential as somewhere with frequent sunny clear skies like the Denver area, for example.


  4. walta100 | | #5

    Thanks for spotting my error I did transpose 2 digits.

    1400 x .9 = 1260 gallons of propane used for heating. 1260 x 91,330 = 115075800 BTU burned. 115075800 x.8 = 92060640 BTUs extracted. 9206040 \ 3412 = 26981 kW

    26981 kW per year for heating


  5. minpix | | #6

    Thanks to Walter, Bill, Akos & Jon for your help on this.
    I'll take the 26981 kW figure to local solar pros to figure out a panel count.
    Much obliged!

  6. Expert Member
    Dana Dorsett | | #7

    From a total lifecycle cost point of view it will be better to run the radiant off a floor thermostat, but use heat pumps to control the room temperatures.

    Better yet, mothball the hydronic system and do all the space heating with heat pumps, which would reduce the total size of the PV array needed by half or more.

Log in or create an account to post an answer.


Recent Questions and Replies

  • |
  • |
  • |
  • |