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how to set air infiltration under 50 pa pressure

asked 2017-09-10 08:26:33 -0500

justin gravatar image

updated 2017-09-11 10:51:55 -0500

I am doing a passive house model, the required airtightness is less than 0.6 ACH at 50 pa difference. In Opensutdio, in space infiltration design flow rate rates, i can set 0.6 air changes per hour but the question is how to set 50 Pascal pressure difference?

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answered 2017-09-11 12:53:58 -0500

You will need to convert the 0.6ACH50 value to a value at "natural pressure" conditions.

There are several ways to do this, of varying speed and accuracy.

What is typically done in residential calculations is to use a correction factor, known as an "n-factor" or "LBL factor" (after Lawerence Berkeley Lab, where it was developed) to convert your air change rate at 50 pascals (ACH50) to an air change rate at natural pressure (ACHnat).

ACHnat = ACH50/n

The n-factor comes from a series of corrections, based on climate zone (temperature influence), the number of stories (stack effect influence), and how exposed the building is (wind influence). Here is a table for looking up the n-factor for U.S. locations:

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If you would like a more rigorous approach, you can derive the factor for your application by doing an hourly calculation with local weather data, similar to how the n-factors were determined above, or do a full airflow-network model in CONTAM or related pressure-network software. Consult the ASHRAE Handbook of Fundamentals chapter on Infiltration and Ventilation for equations on residential infiltration (16.24 in ASHRAE HOF 2017).

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Thanks for your comment! So if the n-factor is 10, then the air infiltration rate is 0.06 ACHnat?

justin gravatar imagejustin ( 2017-09-12 07:28:48 -0500 )edit

That is correct. If you have multiple spaces in the model, I would suggest calculating the total infiltration rate (cfm or m3/s), then dividing that by the exterior facade area (walls+roof) and assigning that in your model as a flow rate per exterior facade area. That way, infiltration is modeled in spaces proportional to their exterior surface area.

mdahlhausen gravatar imagemdahlhausen ( 2017-09-12 12:27:47 -0500 )edit

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Asked: 2017-09-10 08:26:33 -0500

Seen: 57 times

Last updated: Sep 11