Question-and-Answer Resource for the Building Energy Modeling Community
Get started with the Help page
Ask Your Question
5

Is there a way to convert ZoneInfiltration:DesignFlowRate to an equivalent AFN object?

asked 2020-06-25 23:29:00 -0600

saeranv's avatar

updated 2020-06-25 23:31:38 -0600

In EnergyPlus, the ZoneInfiltration objects are not calculated when using the AirflowNetwork (AFN) for the MultiZoneWithoutDistribution control (amongst others). In a previous post[1] it's been suggested that the way to model infiltration in the AFN method, based on the ZoneInfiltration object, is to replace the infiltration objects with enough AFN leakage to get the equivalent infiltration amount of the ZoneInfiltration object. Is there a calculation to figure out this type of conversion out mathematically?

For context, I'm working on a software project where our zones all contain ZoneInfiltration:DesignFlowRate objects, using the Flow/ExteriorWallArea calculation method[2]. So we have the volumetric flow rate m3/s-m2, schedule, constant, temperature and velocity coefficients already defined. We want to include an option to use the AFN for mixed-mode, or purely passive cooling. When this translation is done, we'd like the infiltration rates to be equivalent (or somehow logically translated) from our ZoneInfiltration objects.

Does a formula/look-up table exist to convert the DesignFlowRate to an AFN leakage component, like the AFN:MultiZone:Surface:Crack or AFN:MultiZone:EffectiveLeakageArea objects?

[1] https://unmethours.com/question/39416...

[2] https://bigladdersoftware.com/epx/doc...

edit retag flag offensive close merge delete

Comments

2

Assuming constant value for exterior air infiltration is already a way to model the phenomenon. I guess that the values you are using are derived from an "airtightness indicator" such as N50 (Vol/h @50Pa) or Q4Pa (m3/h.m²@4Pa) for example. If yes, you could use it to compute and create EffectiveLeakageArea objects.

BaptisteDE's avatar BaptisteDE  ( 2020-06-26 06:08:44 -0600 )edit

Good to know, thank you. At this point I'm trying to derive the Crack factor, but maybe in the future things might change to the ELA.

saeranv's avatar saeranv  ( 2020-06-26 17:28:29 -0600 )edit

1 Answer

Sort by » oldest newest most voted
3

answered 2020-06-26 08:08:51 -0600

I've used spreadsheets do do this in the past, but I can't seem to find one right now. But the calculation isn't to difficult to do by hand. In AirflowNetwork and similar models, the flow between two nodes in the network is computed as

mass_flow = f(pressure_difference)

where the function f is determined by the input fields of the particular object. Given enough information (and probably an assumption or two), you can calculate the needed fields. AFN:MultiZone:Surface:Crack uses a power law formulation [1] and a specified reference state, so pick a reference temperature and pressure difference that makes sense and solve

mass_flow = C_Q * pressure_difference^n

for C_Q. For this one, the easy thing to do would be to use n=0.65. That's basically it, but do be careful about the crack factor that comes from the surface that the crack is attached to. For other elements, the procedure is relatively similar, but in other cases it may be harder to make sure that the right reference state is used. We've been adding Python scripts to automate things like this, if that's something you'd be interested in seeing please put in a feature request on the EnergyPlus GitHub repo.

[1] [https://bigladdersoftware.com/epx/doc...

edit flag offensive delete link more

Comments

Thanks for the detailed response Jason! Good to know that it can be done. I'll try and work this out on my own. I'll try and work out the method with the rest of the Ladybug team, but if there's a lot of uncertainty about assumptions or methods I think we'll request the Python script.

saeranv's avatar saeranv  ( 2020-06-26 17:24:47 -0600 )edit

I just did it mathematically like this:

m ̇=C_Q×(ΔP)^n ↔ C_Q=ρ_air(@NTP)×V ̇/(ΔP)^n ↔ C_Q=ρ_air(@NTP)×ACH×V_building/(3600×(ΔP)^n)

I want to respect Passive House's maximum rate of 0.6 ach (at a ΔP of 50 Pa), thus chose 0.2 ach (this is for my dissertation). ρ_air at Normal Temperature and Pressure conditions (the 20℃ and 101325 Pa [1 atm] that are in the Reference by default) is 1.2401 kg/m3, I used the default value of n (0.65), and obtained C_Q=0.00245 kg/s.Pa^n.

@Jason DeGraw, is this way correct, please? If it is, you can use it @saeranv

Mafalda Correia's avatar Mafalda Correia  ( 2020-09-04 03:16:56 -0600 )edit
1

@Mafalda Correia, that looks about right. Keep in mind that if the conditions you're modeling are different than the reference conditions, the flow that you get may be different. So you may want to choose reference conditions specifically for your model

Jason DeGraw's avatar Jason DeGraw  ( 2020-09-04 08:44:23 -0600 )edit

@Jason DeGraw, thanks. I'm new at this, and the house I created is a virtual hypothesis for my dissertation. About the reference conditions: I think the 20ºC seems about right for the mediterranean (warm temperate) climate I'm designing for (south of France), and I want the house to be mostly passive (I'm going to use EMS for hybrid ventilation with an Ideal Loads HVAC system). I now changed the n to 0.8 because I want it to be very airtight. I'm going to do a ratio of external surface areas to assess each zone's C_Q. Do you think the 20ºC, 0.2 ach at 50 Pa, and n as 0.8 are adequate, please?

Mafalda Correia's avatar Mafalda Correia  ( 2020-09-05 12:00:27 -0600 )edit

@Mafalda Correia You are welcome. I'd check the average temperature in your EPW file and take into account the indoor temperature as well (average temperature is used in places). As long as it is close, it shouldn't make a huge difference, but it's always good to know where variations come from. For a more airtight building, you're better off lowering the flow coefficient (that you calculated using 0.2 ach). The exponent in the power law is generally in the interval [0.5, 0.65], going outside of that will be harder to defend than lowering the flow coefficient. I'd leave that as 0.65.

Jason DeGraw's avatar Jason DeGraw  ( 2020-09-06 18:16:51 -0600 )edit

Your Answer

Please start posting anonymously - your entry will be published after you log in or create a new account.

Add Answer

Training Workshops

Careers

Question Tools

3 followers

Stats

Asked: 2020-06-25 23:29:00 -0600

Seen: 557 times

Last updated: Jun 26 '20