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Infiltration values

WPmichael | Posted in General Questions on

I am putting an excel sheet together to get the individual loads from walls, (windows and doors) and infiltration. So I searched the web for the formula to calculate infiltration. I come across this article.

https://soa.utexas.edu/heat-loss-due-infiltration-using-air-change-method

It states “In winter, use 0.5, 0.85, and 1.3 ACH, respectively, for tight, medium, and loose”

Do these ACH value seem reasonable? I understand there is a blower test to determine the actual value.

Thx
Mike

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Replies

  1. user-723121 | | #1

    Understand the difference between natural air change per hour ACHNat and ACH50. In MN we use ACH50 divided by 17 (I believe 20 in CO) for an approximate natural air change for a given building.

    The age of your house may lend some clues as to the type of construction used and the configuration of the house also plays an important role. Think thermal boundary and "surfaces", some homes are efficient designs (a rectangular rambler with south facing glass) and some are sprawling homes with the same square footage but lots of exterior wall surface area.

    7 or 8 ACH50 is common for older homes, the new code for MN is 3. New construction should have a target of 1.5 ACH50 or under, Passive house is .6 as I understand it. Divide these values by 17 for the approximate ACHNat.

  2. Peter Yost | | #2

    Those ACH numbers are for "natural" air changes per hour; meaning not standardized test pressure air changes per hour, such as 50 Pascals, but what the "natural air change is as estimated from the standardized test ACH50. Building scientists hate NACH, because the pressures and corresponding air changes that buildings see under operating conditions are all over the place, from 0 ach to probably 2 ach during a big wind event or somewhere in between when climate conditions are the driver for air changes.

    As a benchmark, ASHRAE 62.2, the residential whole house ventilation standard, recommends that homes need mechanical ventilation when ACH is 0.35 or less (see https://www.energycodes.gov/sites/default/files/documents/BECP_Buidling%20Energy%20Code%20Resource%20Guide%20Air%20Leakage%20Guide_Sept2011_v00_lores.pdf).

    So those numbers for tight, medium, and loose (0.5, 0.85, 1.3) start out a bit high frankly.

    Moving from standardized test pressure blower door results to ACH or NACH involves invoking the N factor, which takes into account wind speed, climate, building height, etc. (see this resource for using the N factor: http://www.pureenergyaudits.com/docs/blower_door_handout_aci_baltimore.pdf.

    Peter

    1. fourforhome | | #4

      I understand that ACH50 (0.2" w.c.) is approximately the differential caused by a 20 mph wind on a surface and that ACH300 is equivalent to a 50 mph wind.
      In building science, the inference is that a 20 mph wind acting on a whole house gives rise to the ACH50 infiltration rate. I doubt this is the case.
      When I had the blower test done on my house (33,000 ft³, 1.25 ACH50), the leaks were whistling and could be felt with the hand. In a real 20 mph wind, I hear and feel nothing like that.
      When a house interior is depressurized by the blower door fan (-0.2" w.c.), makeup air will be found and while it doesn't come through the solid sheathing material, it will come in through cracks and crevices. The makeup air's path of least resistance is through all the cracks and crevices.
      When a house exterior is subjected to a +0.2" w.c. differential, the path of least resistance is no longer through those same cracks and crevices, but rather by going around the house.
      Is it wrong, then, to assume that a 20 mph wind will actually give rise to the infiltration rate of a -0.2" w.c. vacuum? It would seem so.
      The difference might be compared to the relative efficiencies between the old sailing ships that got pushed along by the wind versus the modern aerofoil sails that create a partial vacuum in front of the sail and get pulled along. The vacuum is much more effective at getting work done.
      How is this difference in effect between a -0.2" (internal) and +0.2" (external) w.c. accounted for in ACHnatural and estimated ACH50 (due to wind, not vacuum)?

  3. Jon_R | | #3

    Note that average ACH (used for seasonal energy use) and peak ACH (perhaps useful for a design load) are quite different. In a high wind, ACH-natural could be equal to ACH@50 pascals.

  4. WPmichael | | #5

    Great info, thanks for your input guys.

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