Blower Door / Envelope Leakage Test Certification Requirement

Residential Properties in Florida Permitted after July 1, 2017 Require Testing

When the Florida Building Code, Energy Conservation Version 5 adopted mandatory air infiltration testing (ie, blower door testing) for new construction homes and residential units, enforcement authorities sought greater clarification on who qualified as an “approved entity” to do the testing.

Sections R402.4.1.2 and R104.5 indicate that a Florida blower door test must be done by an approved third party Energy Auditor or Energy Rater. Both titles require certification – either from the Building Performance Institute (BPI) or Residential Energy Services Network (RESNET) – to qualify.

Two Trails is an approved third party provider for blower door testing.  Give us a call to schedule an appointment.

How is the test conducted?

In Florida, it is very common during a blower door test to operate the fan in the depressurizing mode, meaning that air is drawn, pulled or exhausted out of the house. When the fan operates in this way, it is easy for someone walking around inside the home to feel air entering the building through cracks in its exterior envelope (shell). This infiltration commonly occurs around windows, recessed lights, electrical sockets, air handlers, or any feature that penetrates the exterior walls or roof. Inside air leaving the home through the blower (exhausting) has to be replaced with outside air entering the building elsewhere—what goes out, must come in—so spots where air can be felt entering the building are air leakage sites/sources.

What do the numbers from the test mean?

One measure of a building’s air leakage rate is air changes per hour (ACH), which estimates how many times in one hour the entire volume of air inside the building is exchanged with outside air. To determine a building’s ACH, the blower door performs an airtightness test with the house depressurized (or pressurized) to 50 Pascals. A Pascal (Pa) is a small unit of pressure about equal to the pressure that a pat of butter exerts on a piece of toast. Fifty Pascals is approximately equivalent to a 20 mile-per-hour wind blowing against all surface of the building.

The leakier the house, the higher the number of air changes per hour, the higher the heating and cooling costs, and the greater the potential for moisture, comfort, and health problems. Builders of energy-efficient homes generally strive for less than 5 air changes per hour at 50 Pascals pressure (ACH50). Below summarizes the ranges of air changes per hour for typical homes of different levels of air leakage and energy efficiency.

Typical infiltration rates for home ACH50 (Air changes/hour at 50 Pascals)

New home with special airtight construction and a controlled ventilation system 1.5 – 3.0

Energy-efficient home with continuous air barrier system (based on registered ratings) 3.0 – 5.0

Standard new home (based on registered ratings) 5.0 – 7.0

Standard existing home (based on field data) 7.0 – 10.0

Older, leaky home (based on field data) 10.0 – 5.0

To determine the ACH50, we first determine the amount of air flow, measured in cubic feet per minute (CFM), exhausted out of the home with the home at a pressure of 50 Pa with respect to the outside. To calculate this, air flow readings are taken at several house pressure differences, generally between 15 and 60 Pa. These measurements are then entered in a software program to determine the air flow rate at 50 Pa (referred to as CFM50). Multiplying the resulting CFM50 value by 60 (minutes) and dividing by the conditioned volume of the house (in cubic feet) gives the ACH50. For example, a home that has 2,000 square feet of living area and 8-foot ceilings has a volume of 16,000 cubic feet (2,000 square feet multiplied times 8 feet). If the blower door measures a CFM50 leakage of 1,333 then the equation would be 1,333 cubic feet per minute times 60 minutes divided by the volume of 16,000 cubic feet to equal 5 air changes per hour. The home therefore has an infiltration rate of 5 ACH50 (5 air changes per hour at 50 Pa test pressure).

Given ACH50, a natural infiltration rate (resulting from wind and temperature effects) can be estimated for your home. In Florida’s climate, ACH50 measures can be divided by 40 to yield an expected natural infiltration rate, measured as “ACH” (Cummings, Moyer, and Tooley, 1990). For example, if ACH50 = 10 for a Florida home, then the estimate for its natural infiltration rate would be 10/40, or 0.25 ach. This means that under normal wind and temperature conditions, we would expect about 25% of the house air to be replaced with outdoor air each hour with no mechanical equipment running. Note that this is only an estimate of long-term average infiltration. Actual infiltration will vary considerably based on time of day and changes in wind, temperature, and humidity over the course of the day.

In the example home above, with an ACH50 of 5, the expected natural infiltration rate would be 5 divided by 40, or 0.12 ach. This means that under normal wind and temperature conditions, we would expect about 12% of the house air to be replaced with outdoor air each hour.4

How big are the air leakage sites (“holes”) in my house?

Now that you know the CFM50, which is the air flow through the building’s leak sites at a 50 Pa house pressure differential, we can determine the approximate cumulative size of the holes. The “holes” in this case are areas of air leakage. In other words, you are trying to determine the approximate size of all the air leakage sites added together. To do so, multiply the CFM50 by 0.13. In our example above, where CFM50 was 1,333, what the approximate cumulative hole size would be 1,333 times 0.13, or 173 square inches. This means that if you added all the air leak sites together in the example home you would find 173 square inches (1.18 square feet) of potential areas that need sealing, caulking, fixing, or some other sort of repair to reduce leakage.

Are there any limitations to the blower door test “finding” leaks?

As helpful as the blower door test is in measuring and helping to identify areas of air leakage, it’s important to understand that the blower door test does have limitations with respect to locating all sources of leaks. For instance, sometimes indirect leaks or a combination of very small leaks can exist that cannot be easily identified with a blower door test. In such situations, other tools such as infrared thermometers or cameras (discussed later in this fact sheet) can be used to detect changes in temperature that are indicative of air leakage.

Testing the duct system for leaks?

There are a number of tests that can be performed to determine if duct systems in a home are working efficiently. If ducts are not installed correctly, the mechanical systems can be leaky and/or using more energy than necessary. For example, if ducts are located outside the conditioned area of the home, such as in the attic, and they are leaking, the homeowner may not realize they are paying to heat or cool the air outside their living environment.

We use a tool referred to as a duct airtightness tester that can depressurize or pressurize the duct systems and determine the amount of air leakage. The use of a pressure pan in conjunction with the blower door can isolate where ducts are “leaky”.  This test and/or a “smoke” test can be used to help locate duct leaks once a duct airtightness test determines that sealing would be warranted.