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Infiltration Rate for Internal Zone?

asked 2018-06-17 23:27:14 -0600

Katriel gravatar image

updated 2018-06-18 11:22:17 -0600

Hello guys,

I have this building (auditorium) and have established 2 thermal zones: One of them is the Main Room of the auditorium, with no exterior doors or windows, just one door which links the vestibule. The other zone is the vestibule which has 3 operable glass doors.

I have some doubts at modelling it in OS, I want to know first how much is the cooling load in order to size an hvac system later. However, i do not know what goes for the label "Infiltration load" (SpaceInfiltrationDesignFlowRate) for each thermal zone.

The interior doors are completely open for 15 minutes before each event schedule, and the glass doors are fully open 1 hour during the event schedule. Which calculation method is the suitable one for this building?


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answered 2018-06-19 03:50:15 -0600

Isn't that the million dollar question?

Infiltration is always a tough cookie, it's really more of an art than a science unless you can actually measure stuff (eg: blower door for enveloppe infiltration in a closed state), and even measuring can be challenging especially for larger spaces, and is usually something you do at one point in time rather than on the full range of operating conditions...

What follows below is just my quick(ish) two cents, so take it with a grain of salt.

In the case of your auditorium, you have to consider two different infiltration mechanisms:

  • Regular envelope infiltration: no wall, even without exterior windows or doors, is perfect, it's always permeable. You can find some default infiltration values in the ASHRAE Handbook of Fundamentals, which is generally the source for the infiltration values you'll find in the OpenStudio Library Space Types, eg for 90.1-2013 Secondary School Auditorium, that'd be 0.0446 CFM/ft², though that probably includes some windows/doors factored in. If you don't know better, you can always use this type of default value. Otherwise you can estimate and enter an equivalent Air Change per Hour (ACH) number. Note that there is also interaction between infiltration mechanisms and HVAC operation (eg: if your space is positively pressurized by your HVAC system you should expect (much) less infiltration as a result).

  • Doors opened before and after the event.

    • When both the interior and exterior doors are opened
      • You could just estimate a higher ACH (eg: 2.0 ACH) for both the main auditorium and the vestibule.
      • Or if you want to be more accurate, you can perhaps consider that the flow is caused by Wind Only. ASHRAE Handbook of Fundamentals 2013 (or 2009, or 2013-4*n more generally) has a chapter 16 "Ventilation and Infiltration", in which you'll find equation (37) to calculate flow through an opening as a function of wind speed and with an "effectiveness of openings" coefficient you can calculate: the weather file (EPW) you use has both wind speed and wind direction. (There is also equation (38) for stack effect related infiltration, but I don't think it applies much to your vestibule...). Here's equation (37) in IP units:

$$Q = 88.0 \cdot C_v \cdot A \cdot U$$


  • $Q$ = airflow rate, cfm
  • $C_v$ = effectiveness of openings ($C_v$ is assumed to be 0.5 to 0.6 for perpendicular winds and 0.25 to 0.35 for diagonal winds)
  • $A$ = free area of inlet openings, ft²
  • $U$ = wind speed, mph
  • $88.0$ = unit conversion factor

(Note that the chapter has lots of other interesting info, such as infiltration from automatic doors, through an air curtain, etc, as well as examples.)

  • When only the exterior door are opened: I think it's probably fine to just use the regular envelope infiltration above for the main auditorium, but for your vestibule you should jack up the infiltration rate at this point by calculating it ...
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@Neal Kruis: there's still that old bug where mathjax breaks down once you click on "More..." (and just worked once I posted a comment, but a refresh of the page crashes it again)

Julien Marrec gravatar imageJulien Marrec ( 2018-06-19 13:48:58 -0600 )edit

Thanks, @Julien Marrec. I'll check up on this issue.

Neal Kruis gravatar imageNeal Kruis ( 2018-06-20 09:27:05 -0600 )edit

@Julien Marrec, Thanks for the explanation. When you refer to "... 0.0446 CFM/ft²" I found in this page: , page 8 : does it say the same? I'll try to adjust those different values of ACH for each zone based on the building operation schedule.

Katriel gravatar imageKatriel ( 2018-06-20 14:55:44 -0600 )edit

Page 7, Modified App G 2007: "Designed leakage of 0.4 CFM/SF at 75 Pa to be modeled at 0.045 CFM/SF". 0.045 is about equal to 0.0446 indeed (rounding).

Julien Marrec gravatar imageJulien Marrec ( 2018-06-21 04:36:58 -0600 )edit

@Julien Marrec the "More..." issue should be fixed now.

Neal Kruis gravatar imageNeal Kruis ( 2018-07-18 11:31:02 -0600 )edit

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Asked: 2018-06-17 23:27:14 -0600

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Last updated: Jun 19 '18