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ACH50 infiltration heat loss

Lance Peters | Posted in Energy Efficiency and Durability on

CZ 6A, Ottawa ON

So I’m looking into infiltration heat load calculations. The most popular formula looks like this:

ACH50 x volume (cubic feet) x Delta T (deg F) x 0.018 = Load (Btu/h)

So treated as one large room per floor, a 3000 sqft 2 story home with 9ft ceilings and 12″ floor thickness we’re looking at about 60,000 cuft of volume. Let’s assume a pretty tight house at 1 ACH50 and a design temp of -10F for a Delta T of 80F:

1 x 60,000 x 80 x 0.018 = 86,400 Btu/h

Am I doing something wrong? This doesn’t even include the volume of the basement, and at 1 ACH50 it’s hardly an inefficient model.

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Replies

  1. GBA Editor
    Martin Holladay | | #1

    Lance,
    Don't confuse ACH50 -- air changes per hour when a house is depressurized to 50 pascals with a blower door -- with ACH(nat) -- air changes per hour under typical average conditions of wind and delta-T.

    In your climate zone, ACH(nat) = ACH50 divided by 17 (or perhaps divided by 20).

    ACH(nat) is an estimate. Actual infiltration rates will vary widely, based on wind and delta-T.

  2. Jon R | | #2

    Consider what is more appropriate - ACH(natural-peak) or ACH(natural-average). Or assuming that the coldest days aren't likely to be the windiest days, something in between. Windy days can easily reduce the factor from 17-20 to 2-3.

  3. Lance Peters | | #3

    Thanks Martin/Jon for correcting my newbie mistake/misunderstanding!

    So using the 1/17 to 1/20 estimate Martin provides, my ACH(nat) load should be around 4320-5080 Btu/h, a much more believable result!

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