# Estimate kWh electricity needed to heat new addition?

| Posted in GBA Pro Help on

I am planning to expand an existing room in my house. The room is currently on a separate heating zone (oil heat, forced hot water). I have solar panels and generate an average 1700 kWh electricity per year more than I use (this is a 3 year average). So I am thinking of heating the remodeled room with electric baseboard, hoping that I can essentially heat it for free with the excess kilowatts that I generate. But I would like to project how much electricity will be required before making a decision. Can this be estimated?

The room will be 13′ x 17′ (221 square feet).
Ceiling will slant up from 7′ to 15′ so total volume is about 2400 cubic feet.
Total 125 square feet of double-pane glass (two sliding doors plus windows), facing NE and NW.
Closed cell icynene spray foam insulation.
Located near Boston, MA.

Can someone estimate how many kWh per year would be required to heat this? Thank you in advance.

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### Replies

1. GBA Editor
| | #1

User-6963772,
First of all, can you tell us your name?

Q. "Can someone estimate how many kWh per year would be required to heat this?"

A. Yes. These days, the calculation would usually be made using software like RemRate. I suggest that you hire a home performance contractor to help you.

For a back-of-the-envelope calculation, 1,700 kWh is worth about \$204. Your heating season is about 6 months long. Do you think that you can heat this room with electric resistance heat for about \$34 per month? I'm guessing that you'll need more electricity.

2. | | #2

What's the R-value and sf of the walls, floor, ceiling and glass? Do a simple load calc. on your room to get the heat loss. Typical electric space heater is 1,500 watts (1.5 kW and 1.5 kWh per hour) and provides about 5,100 BTUs/ hr.

3. GBA Editor
| | #3

User-6963772,
On second thought, it might work.
Average space heating used in the Northeast (2009 RECS data) for homes between 2,000 and 2,500 sf was 64.9 million BTU.
Assuming an average house size of 2,250 sf, that's 28,844 BTU per square foot.
That's 1,867 kWh of electricity.

4. | | #4

Also consider the whole house big picture. Perhaps a dual head heat pump, allowing your excess electricity to heat the addition and reduce oil usage.

5. Expert Member
| | #5

Heat load isn't proportional to the square footage, so these are REALLY crude estimates, and you have a TON of glazed area. The U-factor of the 125 square feet of glass really matters, and are probably dominating the heat load of that addition, which has something like 3x the typical window/floor area ratio.

IIRC in MA anything in excess of a 15% window/floor ratio means it requires higher than code-min windows to meet code, and you're at ~57%, thuse125 square feet of window for a 221 square foot room would be a code violation unless they are well above code minimum window performance, They may even need to be triple-panes to make it. If double-panes a double low-E argon fill can get you to the low to mid 20s for a U-factor.

6. | | #6

My name is Les and I'm the guy who asked the original question. I am new to this site and naive about a lot of the content and terminology so a lot of the discussion is beyond me. I apologize for my ignorance. Thanks for all your responses.

Windows will be Anderson 200 series and the two sliding doors Anderson Narrowline.

Thanks.

Martin, your rough calculation is encouraging. The projected kWh is almost exactly the excess of what I generate.
.

7. GBA Editor
| | #7

Les,
My back-of-the-envelope pencil scratching may be encouraging, but Dana is absolutely right: You need to perform some real calculations (using Manual J and other software) to determine (a) your design heat load (so that you can size your electric baseboard units) and (b) your annual heating energy use.

8. Expert Member
| | #8

The Anderson 200 series is mostly code-min-ish U-factors, unless you specify the "Low-E PassiveSun w/Heatlok glazing, which I believe includes a second low-E hard coating on surface #4 (similar to the Cardinal double low-E glass), yielding U0.26.

The other glazing options are all U0.28- U0.32, so if you don't know for sure, use U0.30 as your guesstimate.

At a difference of 60F (10F outside, 70F inside, about right for the design temperature a "near Boston" location) under design heating conditions would be 125 square feet of window loses:

U0.30 x 125' x 60F= 2250 BTU/hr, or U0.30 x 125' x 24 hours= 900 BTU per heating degree-day.

That's 10BTU/hr of heat load per square foot of floor area just for your GLASS!

A tight code minimum 2x6 house with a normal glazing fraction would have a heat load of ~10-12 BTU/hr per square foot @ +10F outside for the WHOLE HOUSE(!). That would be including infiltration & ventilation, wall losses, ceiling losses, etc.

A typical Boston year sees ~5600 HDD, so that's 900 x 5600= 5 MMBTU just for the glass losses.

But that much glass also offers a SUBSTANTIAL amount of solar gain, at it will be impossible to estimate the annual heating & cooling energy use on HDD or CDD alone. It needs to be simulate with a better tool than an IBR spreadsheet or Manual-J, and take into account the SHGC of the windows, the shading factors & orientation, roof overhang, and location etc.. BeOpt would be a reasonable tool for getting a handle on that.

9. | | #9

Thanks again for all your responses. I have tried using the online calculators on loadcalc.net and builditsolar.com. There is still a lot I don't understand but I gave it a shot. Loadcalc.net gives what it calls "Total Btu's Heating" (per hour???) and the total is 7950. Builditsolar gives 5568 BTU/hr or 11.5 million BTU/year. Some questions:

1. Is the figure from Loadcalc indeed BTUs/hour? It seems in the same order of magnitude as what Builditsolar gives and for many of the categories (ceiling, floor, walls, etc) the two actually agree pretty closely. So if I was more confident in my interpretation, I can look at where they dont agree very well and see if my assumptions were very different and maybe fine-tune the calculations a little more.

2. Assuming I am on the right track so far, the biggest disagreements seem to be in the categories of glass and infiltration. As for glass I think I can figure out how my assumptions disagree. One site asks for single vs double vs triple pane and whether or not there are shades, whereas the other just asks for an R-value. I can research that some more. But have no idea what parts of my input are going into the calculation of infiltration. Any words of wisdom???

3. Builditsolar reports the total yearly cost (in dollars) and using this and the assumed cost per kWh I can back-calculate kWh per year needed for heat, which was my original question. I come up with about 2400 kWh/year. And multiplying this by the ratio of BTUs estimated from the two sites I can come up with an estimate kWh/year from the other site. I guess my question is, is there a flaw in this thinking?

Finally, I realize that I shouldn't expect these two online calculators to agree exactly, and that neither may be as accurate as professional software. But a ball-park estimate that I have some confidence in is all I am looking for at this point.

Thanks,
Les

10. Expert Member
| | #10

Yes, loadcalc and the IBR approach taken by BIS are in units of BTU/hr.

The U-factors the walls in the BIS calculator may be a bit optimistic, and the U-factors of the windows in loadcalc perhaps a bit pessimistic, but neither are asking for the actual U-factors or SHGC ratings of the windows. This matters quite a bit in this case, since even NW and NE facing windows have real solar gain, even during the heating season (particularly when there is snow cover.)

They are only calculating peak load, then crudely modeling energy use base on that load as if the sun were never going to shine through those windows. loadcalc uses the solar gains when calculating peak cooling load, but not for calculating the heating energy use. Expect the actual use to be lower. But even if it's only 2000 kwh/year, that power still isn't really "free", and leveraging it with a heat pump sized for whatever loadcalc came up with for a cooling load would make some sense. A half-ton or 3/4 ton PTHP would be able to cover the heat load at half the power use of baseboards, a half-ton or 3/4 ton mini-split (while much more expensive than a PTHP) would use 1/3 of the power.

Instead of just burning power in baseboard to get rid of a surplus, consider getting a battery-electric vehicle or plug-in hybrid, and put the excess to a better use than raw un-leveraged BTUs in resistance heaters.

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