[EnergyPlus_Support] Re: cop and iplv values of a chiller, also i have the part load performance

["andresdagm" <andresdagm@xxxxxxxxx>]

Hi, thank you
it helped me, but i see somethnig rare in my model, my chiller consume more that i thinked, i have tried with two diferents model, maybe the problem is the part load curve, i have seen in your spreadsheet:
"IF THE CHILLER MODEL HAS 0.2 AS MINIMUM PART LOAD RATIO,THE SIMULATION WILL RIDE THE "EIR_s" CURVE UNTIL PART LOAD RATIO IS 0.2 AND THEN "
i dont know how to model two curves in one chiller, 

Gracias
--- In EnergyPlus_Support@xxxxxxxxxxxxxxx, Nacho Gomis <nachogo27@...> wrote:
>
> Hi,
> You can read the documentation,the engineering reference pdf file.It has
> the chiller models very well explained.
> I have also uploaded an spreadsheet to the yahoo group which leads you to
> obtain the three curves after providing the performance chiller data values.
> I hope this can help you.
> Ignacio Gomis
> 
> ---------- Forwarded message ----------
> From: andresdagm
> Date: Tuesday, 31 July 2012
> Subject: [EnergyPlus_Support] cop and iplv values of a chiller, also i have
> the part load performance
> To: EnergyPlus_Support@xxxxxxxxxxxxxxx
> 
> 
> 
> 
> dear,
> Good day,
> I have the cop and iplv values of a chiller, also i have the part load
> performance values of
> EEFT Evaporator entering fluid temperature
> ELFT Evaporator leaving fluid temperature
> CEFT entering condenser water temperature
> CLFT leaving condenser water temperature
> CAP (ton)
> PCT Power
> kw/TR
> for each part load(15,20,30...100%)
> 
> I dont know how to set it or how to transform it to can set this values in
> energy+, i think that the program could generate a curve with only COP and
> IPLv but i havent cant find that. Also i have tried to use Chiller:EIR and
> the EIRPLR curve is:
> 
>   EIRFPLR   y = 1.9927x2 - 1.9632x + 1.0078
> R² = 0.9499
> 
> i dont know if it is correct, also i dont know set up the other 2 curves
> because both are for full load, well, i think that because there ae onl
> function of temperatures.
> 
> thank you in advance.
>




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