[Equest-users] Chiller Curves

[Rob Hudson]

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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