ACH50 infiltration heat loss
lance_p
| 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.
GBA Detail Library
A collection of one thousand construction details organized by climate and house part
Search and download construction details
Replies
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.
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.
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!