Oh right, my mistake. You generally want to double the area of internal mass’s area to model both the front and back sides of it.
From: EnergyPlus_Support@xxxxxxxxxxxxxxx [mailto:EnergyPlus_Support@xxxxxxxxxxxxxxx] On Behalf Of Vinay Devanathan
Sent: Friday, July 22, 2011 2:59 PM
To: EnergyPlus_Support@xxxxxxxxxxxxxxx
Subject: Re: [EnergyPlus_Support] Internal Mass object calculation
Got it. From Eric's reply, checked the floor area and internal mass surface area, the ratio is two times as mentioned.
Thanks, Eric & Brent!
On Fri, Jul 22, 2011 at 4:55 PM, Griffith, Brent <brent.griffith@xxxxxxxx> wrote:
If I remember right, the algorithm assumes internal mass is six inches thick of wood with same area as the zone’s floor area.
From: EnergyPlus_Support@xxxxxxxxxxxxxxx [mailto:EnergyPlus_Support@xxxxxxxxxxxxxxx] On Behalf Of Vinay Devanathan
Sent: Friday, July 22, 2011 2:49 PM
To: EnergyPlus_Support@xxxxxxxxxxxxxxx
Subject: [EnergyPlus_Support] Internal Mass object calculation
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