[Equest-users] Chiller Curves

[Carol Gardner]

Hi Rob,

It looks like you all ready solved your problem, which is good, but I'm
going to send you a comment anyway.

When I create a curve, I always use "raw data points" instead of
"coefficients". The reason I do is that I can always obtain them and I don't
have to worry about making a calculation mistake because eQUEST takes the
raw data points and calculates the coefficients for me. If I want to see
what they are, I can look in the .bdl file.

The only trick to raw data points is that for everything except
refrigeration compressors the dependent, in this case PLR, needs to be
normalized to either the ARI design, or the actual design if it is
different.

So for your chiller curve you would input: Independent    Dependent

1.0                     1.0

0.75                   0.7356

0.50                   0.6643

0.25                   0.582

Anyway, that's my 2 cents.

Carol

On Wed, Jul 20, 2011 at 10:24 AM, Rob Hudson <rdh4176 at gmail.com> wrote:

> Prior to your response, i got a little impatient and found some of your
> past articles in the archives after a lot of digging.
>
> I found out that the EIR equation is actually a % of the EIR that needs to
> be taken into account.  So the equation should output a value from 1 to 0 to
> multiply the design rated EIR by to get the usage at that part load.  there
> was a method that i found that, when the manufacturer doesn't provide you
> with enough to populate all three of the custom curves, you can simply use
> the EIR f(PLR) curve, which i did and got the results i expected to see.
>
> Thanks for your response, both past and present.
>
> Rob
>
>
> On Wed, Jul 20, 2011 at 1:06 PM, Nick Caton <ncaton at smithboucher.com>wrote:
>
>>  Hi Rob!****
>>
>> ** **
>>
>> Well to start, let’s not assume every chiller out there MUST be more
>> efficient than the default curve at every point.  Better to assume they can
>> and should differ.****
>>
>> ** **
>>
>> With that disclaimer out of the way,  you have noted your coefficients
>> generated for your custom PLR curve should produce a lower correction factor
>> at 25% (confirmed in the Var 9 column you attached).  ****
>>
>> ** **
>>
>> What I think you’re missing is that there is indeed more than one
>> correction factor being calculated every hour, and more than one curve in
>> play to produce those factors that you need to consider.  The extra column
>> you chose to include in your spreadsheet hints at this (Var 10).****
>>
>> ** **
>>
>> Recommended DOE2 reading that should fill in these specific gaps for you:
>> Volume 2: Dictionary > HVAC Components > CHILLER > Chiller Energy
>> Consumption****
>>
>> ** **
>>
>> There much further discussion on this topic in the archives as well, if
>> you just can’t get enough ;).****
>>
>> ** **
>>
>> ~Nick****
>>
>> ** **
>>
>> [image: cid:489575314 at 22072009-0ABB]**
>>
>> * *
>>
>> *NICK CATON, P.E.***
>>
>> SENIOR ENGINEER****
>>
>> ** **
>>
>> Smith & Boucher Engineers****
>>
>> 25501 west valley parkway, suite 200****
>>
>> olathe, ks 66061****
>>
>> direct 913.344.0036****
>>
>> fax 913.345.0617****
>>
>> www.smithboucher.com* *****
>>
>> ** **
>>
>> *From:* equest-users-bounces at lists.onebuilding.org [mailto:
>> equest-users-bounces at lists.onebuilding.org] *On Behalf Of *Rob Hudson
>> *Sent:* Wednesday, July 20, 2011 6:23 AM
>> *To:* equest-users at lists.onebuilding.org
>> *Subject:* [Equest-users] Chiller Curves****
>>
>> ** **
>>
>> I have a new Chiller Curve question/situation.  I have created a custom
>> chiller curve for the EIR = f(PLR) as a cubic function using the following
>> data inputs:****
>>
>> ** **
>>
>> 1         .3313****
>>
>> .75      .2437****
>>
>> .5        .2201****
>>
>> .25      .1928****
>>
>> ** **
>>
>> From my understanding, the chiller curve seems correct.  However, when i
>> create a hourly report summary for the Corrected Chiller EIR and compare it
>> to the default EIR equation for this entry, i get 0.034 for the default and
>> 0.072 for my curve at 25% ratio.  Why is it that the default is better than
>> my curve?  I have looked at the equation for the default entry and got the
>> following equations:****
>>
>> ** **
>>
>> EIR = a + b x = c x^2****
>>
>> a = .04812248****
>>
>> b = .69573420****
>>
>> c = .23493889****
>>
>> ** **
>>
>> at 25% this equation results in 0.2367, and my equation gets 0.1928****
>>
>> ** **
>>
>> Is there any other calculations going on in the background that is causing
>> some problems?  I have attached an excel snapshot of the different EIR and
>> the corrected EIR values for two different chillers in my program.  Neither
>> is currently using the default EIR = f(PLR) curves, they are using custom
>> curves.****
>>
>> ** **
>>
>> Thanks in advance,
>>
>> --
>> Rob Hudson****
>>
>
>
>
> --
> Rob Hudson
>
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>


-- 
Carol Gardner PE
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