Is my house efficient?

josephny

I have a 4500 sq ft house (3 stories) in climate zone 6A, very well insulated but lots of (good quality) windows. 

In both November and December the three LG heat pumps together used about 1800 kWh of electric to keep the house an average of 69 or 70 degrees. 

Is my understanding correct that I multiply 1800 kWh by 3400 kWh/Btu to arrive at a total Btu use of 6,120,000. Divide they by 30 days, and that by 24 hours and get 8,500 Btu/hr. 

Does this tell me anything useful?

Is this my heating load (on average across these 2 months)?

Is this a lot or a little amount of energy to heat this house?

Am I completely misunderstanding?

Hot_water_fan

You’re pretty close!

1. It’s 3400 Btu/kWh, not kWh/Btu. Probably just a typo. 

2. You’re assuming a COP = 1, which is true for electric resistance but isn’t true (on average) for heat pumps. So your 1800kwh INPUT is probably around 3 x 1800kwh OUTPUT. Since you pay for input, it’s good to know that too, but your house is heated by output. 

3. It’s kind of useful to know average hourly heat loss, but you size based on design temp. So that would be how many kWh on the coldest day, not average day. If you see “20btu/sqft” on the Wall, that’s based on the max heat loss, not an average.  

4. Find the heating degree days for your area for those months and divide by that. For example, I’m at 1.15kwh per HDD (base 60F). So knowing a winter’s average HDD, you can figure about how many kWh you’ll use annually.

I think based on those figures, your house and system is pretty efficient!  

josephny

Thank you for the explanation. Working on understanding it.

Typo: Btu/kWh -- thank you.

I have a basic-enough understanding of COP, and the more data and analysis, the better, but wouldn't I just be guessing about my actual COP by doing the analysis this way? That is, without know the actual energy in and energy out, how could I know the COP? If true, then I might as well ignore the role of COP in this (it could be an average of 1.5 or 3.5, and varying by outside temp from 0 to 4)?

So I looked at:

www.degreedays.net

and discovered that the coldest day in the past 2 months (10/22/2023 through 12/24/2023) was 12/22/23 with 35.7 HDD with a base of 60 *F.

On the same day, I used 103 kWh of electric to heat the home to about 68 *F

Does this mean that on 12/23/2023 I used 2.88 (103/35.7) kWh/HDD?

Picking another data randomly, 11/28/2023 had 29.6 HDD and used 97 kWh -- 3.28 kWh/HDD

And I doing this incorrectly?

Is 3 kWh/HDD high?

Hot_water_fan

Correct, you don’t know your COP. What came before the heat pump? You can guess at it by comparing the BTUs from before to now. Or just use a number like 2.5 - 3. 

I think 3kwh/HDD is great for a house that size. Over a winter with maybe 4000hdd that’s just 12000kwh. That’s about 1000 gallons of oil, assuming a COP = 3.