RE: [EnergyPlus_Support] Re: Window blinds - conductance of shading layer in heat balance equations

Right. It is as if the blind is “closed”.

From: EnergyPlus_Support@xxxxxxxxxxxxxxx [mailto:EnergyPlus_Support@xxxxxxxxxxxxxxx] On Behalf Of tboomgaert
Sent: Friday, April 24, 2009 4:13 AM
To: EnergyPlus_Support@xxxxxxxxxxxxxxx
Subject: [EnergyPlus_Support] Re: Window blinds - conductance of shading layer in heat balance equations

Ok, I think I just misunderstood the concept. So the point is then that the left and right surface temperature of the shading layer are meant to be equivalent to the bottom and top surface temperature of a slat?

--- In EnergyPlus_Support@yahoogroups.com, "Griffith, Brent" <brent.griffith@...> wrote:
>
> I like your idea, but that is not what the program is doing. The blind
> model is only for relatively normal slat configurations with low
> conductance.
>
>
>
>
>
> ________________________________
>
> From: EnergyPlus_Support@yahoogroups.com
> [mailto:EnergyPlus_Support@yahoogroups.com] On Behalf Of tboomgaert
> Sent: Thursday, April 23, 2009 3:49 AM
> To: EnergyPlus_Support@yahoogroups.com
> Subject: [EnergyPlus_Support] Re: Window blinds - conductance of shading
> layer in heat balance equations
>
>
>
>
>
>
>
>
> Thank you for your answer.
>
> I don't understand this completely however. Do you mean that the shading
> layer conductance k is just equal to lambda/t (with lambda the slat
> conductivity and t the slat thickness)? I find it strange that the
> distance between the slats is nowhere included in this, as well as width
> of the slats (and equivalent shading layer).
>
> I was thinking that the conductance would have to be calculated somewhat
> like this:
> k=lambda/((t/(h+t))*w)
> so that:
> - for slats with no thickness, the conductance would be infinite
> (because there is no material)
> - for h=0 (no distance between slats), the conductivity of the shading
> layer would be equal to the slat confuctivity, so that k=lambda/w
>
> Can you tell me if I'm right or not?
>
> Thanks in advance,
> Thomas
>
> --- In EnergyPlus_Support@yahoogroups.com
> <mailto:EnergyPlus_Support%40yahoogroups.com> , "Griffith, Brent"
> <brent.griffith@> wrote:
> >
> > Blind layer conductance is calculated by dividing slat conductivity by
> > slat thickness.
> >
> >
> >
> > ________________________________
> >
> > From: EnergyPlus_Support@yahoogroups.com
> <mailto:EnergyPlus_Support%40yahoogroups.com>
> > [mailto:EnergyPlus_Support@yahoogroups.com
> <mailto:EnergyPlus_Support%40yahoogroups.com> ] On Behalf Of tboomgaert
> > Sent: Friday, April 10, 2009 4:13 AM
> > To: EnergyPlus_Support@yahoogroups.com
> <mailto:EnergyPlus_Support%40yahoogroups.com>
> > Subject: [EnergyPlus_Support] Window blinds - conductance of shading
> > layer in heat balance equations
> >
> >
> >
> >
> >
> >
> >
> >
> > Hello,
> >
> > In the Engineering Reference on p.228-230 (heat balance equations for
> > shading device and adjacent glass), it states that shading devices are
> > treated as a uniform layer with a conductance k_sh.
> >
> > I was wondering how this k_sh is calculated for window blinds (with
> > horizontal slats)? I can't seem to find anything on this in the
> > engineering reference.
> >
> > Thanks in advance!
> > Thomas
> >
>

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