Air leaking into a home (infiltration) or out of a home (exfiltration) happens naturally in every home, new or old. No matter how much air sealing is performed, we just can’t make them completely air tight. I’ve tested some new homes that were very tight, .33 ACH50, (anything under 1 ACH50 is very good) and I’ve also tested many older ones that aren’t so tight, we can use my 1952 Cape as an example, 9.71 ACH50. In this post, I’m going to discuss how to manually calculate the cost of the air leakage and examine what we can do with that number.

To calculate the cost of air leakage, we will first need to perform a blower door test to get the air leakage rate in cubic feet per minute at 50 Pascals (CFM50). Other information that will be needed to calculate the **heating** cost of the air leaks includes the cost of the heating fuel per BTU, the efficiency of the heating system, the heating degree days and n-Factor for the location of the home. To calculate the **cooling** costs of the air leaks, we need cooling degree days (CDD), cost of electricity per kWh, and the seasonal energy efficiency ratio, or SEER of the cooling system will be needed. Sounds complex, but it’s really not that hard.

Let’s start with calculating the cost of the heating fuel per BTU. For those of you who do not know what a BTU is, BTU stands for British Thermal Unit, a measurement of heat, 1 BTU has the equivalent heat output of about one burning match. You’ll need to know the cost and BTU output of the heating fuel for a given volume. For instance, natural gas has a BTU output of approximately 100,000 BTU’s per therm, and in my area, natural gas is around $1.50 per therm. The formula is:

**Fuel cost / BTU’s per unit (gallon, therm, kWh etc…)**

In the case of natural gas, $1.50 / 100,000 BTU’s = $.000015 per BTU.

The heat output for different fuel sources:

**Natural gas-1 therm = 100,000 BTU**

**Propane-1 gallon = 91,500 BTU**

**Fuel Oil – 1 Gallon = 138,500 BTU**

**Electricity-1 kWh = 3413 BTU**

**Wood-1 cord = Approximately 20,000,000 BTU**

Now that you have the heat output for different the different fuel sources, simply use the formula above to calculate the cost of the fuel per BTU based on fuel costs for your area.

For the cooling cost calculation, we simple need the cost of electricity for the home. Electricity costs in my area are $.12/kWh.

Next we need to know the heating degree days (HDD) for the location of the home. Heating degree days is the average temperature for a given day below 65°F. So if the average temperature in a given day is 35°F, there are 30 heating degree days for that day. 65°F – 35°F (average temperature for the day) = 30°F HDD. Add up all the daily heating degree days for a location and you have the annual HDD’s for the formula. You don’t have to calculate all those numbers manually, there are a few websites that will give you the HDD for your location. The site I most often use is https://www.degreedays.net/ Just make sure you get the average yearly number. The HDD for my area over the past year were 8,993. A little below what we typically see, which is closer to 9,500.

Cooling degree days (CDD) will need to be known for the cooling calculation. Calculating the CDD is the same as HDD except 78°F is used, the average daily temperature above 78°F. If we had an average daily temp of 80°F, there would be 2 Cooling Degree days (80°F – 78°F = 2 CCD). My area had a whopping 94 CDD over the past year.

Next up is the n-Factor. The n-Factor (also called the LBL Factor) was developed a few decades ago by the Lawrence Berkeley Laboratory (LBL) as a way to calculate the natural air change rate by using the blower door test results. LBL came up with a map of the US and Canada that uses wind data for the given location, how well shielded the home is, and how many stories the home has. The illustration and chart are used to determine the n-Factor for a given area.

I’m located in the arrowhead of Minnesota, zone 2. We will need to know the number of stories and how well shielded from the wind the home is to get the n-factor number from the table.

Last piece of information is the efficiency of the heating and cooling equipment. This can be found on the EPA’s required tag that comes with the heating system, or in the case of an air conditioner, you may need to look up the manufacturer and model number online to get the SEER. The SEER for most air conditioners will be between 10 and 20, though older units can be less than 10.

Now that we’ve collected all the information, we can input the info into a formula. The formula for calculating the heating cost of air leaks is:

**(26 x HDD x Cost of heating fuel/BTU x CFM50) / (n-Factor x Efficiency of the heating equipment) = Heating cost of the air leaks**

The number 26 is a constant and I believe it has to do with the heat capacity of air.

The formula for cooling is:

**(.026 x CDD x cost of electricity per kWh x CFM50) / (****n-factor x SEER) = cooling cost of air leaks**

We will use the two houses I mentioned earlier to calculated the cost of the air leaks, the tight house at .33 ACH50 and my leaky house, 9.71 ACH50. Let’s start with my home, the heating degree days, as mentioned earlier is 8993. My home blower door tested at 2976 CFM50. I heat with natural gas which the cost per BTU was also calculated earlier, $.000015/ BTU. The home is a story and a half with normal wind shielding, zone 2, the n-Factor we will use is 16.7. And lastly, the heating system efficiency is 96%, (this is the furnace’s AFUE rating, not taken into account is the total efficiency of the system including any ductwork inefficiencies, which would drastically lower this value.) Plug all these numbers into the formula and you get:

**(26 x 8993 HDD x $.000015 x 2976 CFM50) / (16.7 x .96) = $651 per year**

Now we can calculate the cost of the air leaks for the cooling season. Cooling degree days is 94, cost of power is $.12 per kWh, and the SEER of the air conditioning system is 13:

**(.026 x 94 x $.12 x 2976 CFM50) / (16.7 x 13) = $4.02**

So, for my home, I’m spending $655 on increased heating and cooling costs based on the air leakage of the home. This number has nothing to do with heat loss due to low insulation levels, just air leakage.

Now let’s take a look at the very tight home, .33 ACH50, this is the actual blower door test results for a home I recently tested. 91 CFM50 was the air leakage rate. Heating degree days is still 8893. This home was rural and the heating system fuel is propane. The current cost of propane in my area is $1.80/gallon and the heat output is 91,500 BTU, this results in a cost of $.00002 per BTU. The home was also a one and a half story in a normal wind shielded area in the same zone, so the n-Factor remains at 16.7. Lastly, the home is heated using a condensing boiler, 96% efficient. The numbers are very close to my 1952 Cape with the exception of the air tightness level.

**(26 x 8993 HDD x .00002 x 91) / (16.7 x .96) = $26.54**

Cooling calculation using many of the same numbers as my home, except the cooling system is new and has a better SEER rating of 17.

**(.026 x 94 CDD x .$12 x 91 CFM50) / (16.7 x 17) = $.095**

The very tight home’s cost of the air leakage in both heating and cooling is $26.63 per year. That’s a substantial savings over the 1952 Cape.

What can we do with these calculations? Well, let’s say we want to improve the air tightness of a home and put a financial number to the improvements. I plan on cutting the air leakage rate by 50% in my home, spread that savings over a 10 year period and I will save around $3275. Realistically, this would be a good number to invest into the air sealing measures, I would end up with a 10 year payback on my investment. In comparison, it wouldn’t make financial sense to try to improve the air sealing in the already well air sealed home. I can almost guarantee the expense to reduce the air leakage rate of this home by any more than a few CFM would cost substantially more than any savings realized.

The take away, if you are building new, concentrate on air sealing, there are many benefits, energy costs savings is one. If remodeling, conduct some testing of the home, there may be a financial benefit with air sealing in renovation work as well.