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RE: [EnergyPlus_Support] very high U-value

Invert the material conductivity to get the resistance of 1 m thickness

 

Multiply by the material thickness to get the layer resistance

 

Invert the layer resistance to get the U-Value

 

17.04 is the result

 

Ned Lyon
Staff Consultant
 

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Waltham, MA 02453



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From: EnergyPlus_Support@xxxxxxxxxxxxxxx
[mailto:EnergyPlus_Support@xxxxxxxxxxxxxxx] On Behalf Of Simge Andolsun
Sent: Monday, October 29, 2007 12:02 PM
To: EnergyPlus_Support@xxxxxxxxxxxxxxx
Subject: [EnergyPlus_Support] very high U-value

 

Linda Lawrie,

For a simulation study, I defined the name and the layers of the floor
construction as shown in Table 1. I then entered the HF-C5 with the
properties listed in Table 2 into MATERIAL:REGULAR object of EnergyPlus
(Table 1).

Table1. CONSTRUCTION

Name LTFLOOR
Outside Layer HF-C5

Table2. MATERIAL:REGULAR

PROPERTIES HF-C5
Roughness Medium rough
Thickness (m) 0.1015
Conductivity (W/m-K) 1.7296
Density (kg/m3) 2243
Specific Heat (J/kg-K) 837
Absorptance: Thermal 0.9
Absorptance: Solar 0.65
Absorptance: Visible 0.65

I ran the simulation and EnergyPlus calculated the U value of the floor
as
17.04. This value is very high and it does not seem to be right. What do
you think? I would like to know more about how EnergyPlus calculates the
U-values of floors. Where may I get this information from?
I would be glad if you could help me.

ps: the related .idf file is attached to this e-mail.

Thanks in advance for your reply,
Simge Andolsun

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