# Equating heat loss to natural gas usage

I am trying to equate the heat loss from my house to equate with my natural gas usage.

I have a mid-1950s 11,385 sf ranch home in PA, zone 5. I performed a blower door test to measure my air loss, arriving at CFM50 rate of 1512 or an ACH50 rate of 7.97.

Going by the heat loss formula:

** Q = V x ACH x 0.0182 x HDD **** x 24**

Where:

Q = Heat loss

V = Volume

ACH = Air changes per hour at 50 Pa

HDD = Heating Degree Days

Looking at January of 2019, we had 1012.4 HDD for my region.

This would mean

**Q = 11,385 x 7.97 x 0.0182 x 1012.4 x 24 = 40,126 kBtu**

So for the month of January, I had 40,126 kBtu exfiltrate from my house.

Using Energy Star Tech Reference, there are 102.6 kBtu/ccf. This would mean I should be using about 391ccf of natural gas to maintain my 65 degree setpoint.

But I only used 100 ccf in January.

My initial thought was maybe this was due to furnace inefficiencies/unaccounted natural gas use (water heater, etc), but that would mean the ratio should be reversed: I should have used *more* natural gas than is exfiltrating from the house.

I actually did the heat loss calc for my natural gas consumption for the past two years, using excel, and found that over the winter months (Dec-Mar) the ratio is consistently around 4x what it should be. Accounting for water heater use and other odd variances, I am guessing there is a miscalc or missing variable that would lead it 4 times higher than it should be?

Anyone see what I’m missing?

(Thank in advance for any assistance)

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## Replies

The air infiltration, which is only part of the heat load. Conducted heat out of all the exterior surfaces of the house is usually a much larger number. The natural air infiltration rate is much MUCH smaller than when the house is at 50 pascals pressure relative to the outdoors, and estimating the ACH/natural from the ACH/50 measurement is extremely imprecise.

Also where is the correction for the equipment's combustion efficiency? Where is the 99% outside design temperature being factored in?

To run a load calculation based on fuel use, try this approach:

https://www.greenbuildingadvisor.com/article/out-with-the-old-in-with-the-new

If you're maintaining the interior set point at 65F, use base 60F heating degree days for inferring the linear constant.

Also, can I assume the house size is actually 1,385 square feet, and not 11,385 square feet?

The equation was pulled directly from Krigger's "Residential Energy: Cost Savings and Comfort for Existing Buildings" (2014) text. I might have misread the text...I think I was assuming that was giving me the total heat loss, but I see it was just the air infiltration.

Thank you for the article, I will definitely use that and see how it comes out.

The house dimensions are 23 x 33 ft, with 7 ft basement ceilings and 8 ft first floor ceilings. The entirety of both floors make up the conditioned space.

I like to use Btu's per square foot per heating degree day as a heating energy use measurement. I have measured my own house this way for years and it will also show the increasing solar fraction as a slightly lower Btu/sf/hdd number as daylight increases. With your ACH50 number and the volume of the house you should be able to assign an infiltration heat loss number in Btu's.

As Dana says, your formula has two problems:

a) it doesn't account for conduction losses

b) it assumes 50 pascals of pressure, which is way too high for the typical case

A more reasonable guess would be to 1/20 of your calculated value as the air leakage heat loss. Say 2Mbtu, leaving 8.28Mbtu due to conduction. Maybe 5.79 btu/sf/hdd (January, conduction only).