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Why would both frame and divider conductances be 500 W/m2-K?

asked 2017-07-19 21:16:21 -0600

sirYESsir gravatar image

updated 2017-07-19 22:19:32 -0600

I have the 4 objects in my WindowProperty:FrameAndDivider class and I am struggling to understand what each of those are representing, such they are all so similar, but with major differences, specially their conductances. Here they are: (I will highlight the relevant fields alone)

1,                       !- Name
9.5,                     !- Frame Conductance {W/m2-K}
0.5,                     !- Frame Solar Absorptance
0.5,                     !- Frame Visible Absorptance
0.9,                     !- Frame Thermal Hemispherical Emissivity
DividedLite,             !- Divider Type
9.5,                     !- Divider Conductance {W/m2-K}
0.5,                     !- Divider Solar Absorptance
0.5,                     !- Divider Visible Absorptance
0.9,                     !- Divider Thermal Hemispherical Emissivity
0.5,                     !- Outside Reveal Solar Absorptance
0.5;                     !- Inside Reveal Solar Absorptance


2,                       !- Name
500,                     !- Frame Conductance {W/m2-K}
0.3,                     !- Frame Solar Absorptance
0.3,                     !- Frame Visible Absorptance
0.3,                     !- Frame Thermal Hemispherical Emissivity
DividedLite,             !- Divider Type
500,                     !- Divider Conductance {W/m2-K}
0.3,                     !- Divider Solar Absorptance
0.3,                     !- Divider Visible Absorptance
0.3,                     !- Divider Thermal Hemispherical Emissivity
0.3,                     !- Outside Reveal Solar Absorptance
0.3;                     !- Inside Reveal Solar Absorptance

3,                       !- Name
9.5,                     !- Frame Conductance {W/m2-K}
0.5,                     !- Frame Solar Absorptance
0.5,                     !- Frame Visible Absorptance
0.9,                     !- Frame Thermal Hemispherical Emissivity
,                        !- Divider Type
9.5,                     !- Divider Conductance {W/m2-K}
0.5,                     !- Divider Solar Absorptance
0.5,                     !- Divider Visible Absorptance
0.9,                     !- Divider Thermal Hemispherical Emissivity
0.5,                     !- Outside Reveal Solar Absorptance
0.5;                     !- Inside Reveal Solar Absorptance

4,                       !- Name
500,                     !- Frame Conductance {W/m2-K}
0.3,                     !- Frame Solar Absorptance
0.3,                     !- Frame Visible Absorptance
0.3,                     !- Frame Thermal Hemispherical Emissivity
,                        !- Divider Type
500,                     !- Divider Conductance {W/m2-K}
0.3,                     !- Divider Solar Absorptance
0.3,                     !- Divider Visible Absorptance
0.3,                     !- Divider Thermal Hemispherical Emissivity
0.3,                     !- Outside Reveal Solar Absorptance
0.3;                     !- Inside Reveal Solar Absorptance

Does anybody have any idea of what that 500 W/m2.K represent? I thought of eventual infiltrations, but those are already being addressed in the ZoneInfiltration:DesignFlowRate class.

Thanks in advance for any help.

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answered 2017-07-20 04:50:00 -0600

I think Frame Conductance here really is just the conductance of the frame per square meters. Maybe it's just me, but when I see conductance, I usually think about thermal conduction inside a single material, represented by k [W/mK], but since the unit of Frame Conductance here is W/m2K this just probably means, that when overall heat transfer is calculated, e+ calculates it with the heat flux density, instead of heat flux. It is most accurate, if you use a value calculated by Berkeley Lab WINDOW as it said in the InputOutputReference.

In this case the high value of 500 W/m2K may be used, because the frame in question is very thin and made of metal, like an old industrial steel framed window.

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@adambgnr: Thank you for your reply. So if I'd like to adjust this value to consider other materials for the frame (say, wood: U = 3 W/m^2.K), I assume I should know either its material or its thickness. How can I assess any of these?

sirYESsir gravatar imagesirYESsir ( 2017-07-20 06:48:35 -0600 )edit

You can get this data from Berkeley Lab WINDOW, if you model the window there.Be careful with U, it is the heat transfer coefficient, not the conductance in question. While it is true, that confusingly they have the same unit here, U takes into account for the film resistance on the surfaces as well, while Conductance tells, how easily can heat flow inside the material.

adambgnr gravatar imageadambgnr ( 2017-07-20 07:45:37 -0600 )edit

@adambgnr: Isn't your suggestion a solution for my reversed problem, as in getting the data after modelling the window? What I actually need to know is the material that constitutes the frame or its thickness just by knowing the information contained in the e+ object. Other than the info. contained in the objects, I don't know anything else, such I wasn't the one designing the building.

sirYESsir gravatar imagesirYESsir ( 2017-07-20 08:47:01 -0600 )edit

@adambgnr: I was whatsoever trying to run the software Berkeley Lab WINDOW but there doesn't seem to exist any platform to open the .mdb file (it keeps opening it on Microsoft Access). How do you open it?

sirYESsir gravatar imagesirYESsir ( 2017-07-20 09:29:42 -0600 )edit

"What I actually need to know is the material that constitutes the frame or its thickness just by knowing the information contained in the e+ object": I think there are multiple solutions to this problem, since conductivity and thickness are both free variables. Maybe the simplest solution is to ask the designers, or just make an approximation.

adambgnr gravatar imageadambgnr ( 2017-07-20 10:56:12 -0600 )edit

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Asked: 2017-07-19 21:16:21 -0600

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Last updated: Jul 20