[Equest-users] Default Performance Curves

[John T. Forester]

Cliff,

There is no square term.  Here is an example of normalizing the EIRf(PLR) curve for a fan or pump with known kW (or bhp and motor efficiency) at given part-load speeds and at full-load speed.  The data below is fictitious so I don’t recommend using it for a real curve.

PLR

kW

20%

0.98

30%

1.63

40%

2.47

50%

3.56

60%

4.97

70%

6.75

80%

8.98

90%

11.70

100%

15.00


To normalize this data, divide the kW for each part-load value by the full-load kW.  Doing this, your data will look like the table below.  The idea is that when a PLR of 1.0 is entered, the result is 1.0.  This type of curve is used to scale some other input or eQuest-calculated value.

PLR

Normalized kW

20%

0.066

30%

0.109

40%

0.165

50%

0.238

60%

0.331

70%

0.450

80%

0.598

90%

0.780

100%

1.000


Next, you can use MS Excel to plot this and match a 2nd order trendline to it and show the trendline equation.  These are the coefficients you want to enter into a custom performance curve.  Your plot with trendline should look similar to below.  Just remember that eQuest asks for the coefficients in reverse order than how MS Excel plots them.  eQuest will ask for A + Bx + Cx2.  Also pay attention to negative and positive signs.  If the coefficient has a negative sign, be sure to include it in eQuest.  You will see that a 2nd order trendline will fit your data quite nicely due to the characteristics of your system.  For equipment such as variable speed fans on cooling towers or condensing units, the data should more closely match the affinity laws or a 3rd order trendline.  This is because there is relatively little equipment downstream of the fan for the pressure to change at varying velocities.

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John T. Forester, P.E., LEED AP, Mechanical Design Engineer I BVH Integrated Services I 617.658.9008 tel I 617.244.3753 fax I One Gateway Center Suite 506, Newton MA 02458 I www.bvhis.com<http://www.bvhis.com> I Hartford ● New Haven ● Boston
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From: equest-users-bounces at lists.onebuilding.org [mailto:equest-users-bounces at lists.onebuilding.org] On Behalf Of Clifford McDonald
Sent: Thursday, October 14, 2010 9:09 AM
To: Erik Kolderup
Cc: equest-users at lists.onebuilding.org
Subject: Re: [Equest-users] Default Performance Curves

Thanks for the help on this, but I'm still confused. Are you saying the form of the equation is:

PLR = (PLR * (EIRcurrent / EIRfull) ) + (PLR * (EIRcurrent / EIRfull) )^2

?  Then, if the equipment has constant efficiency across the full load range, the EIR/EIR term is one, and you get:

PLR = PLR + PLR^2, which doesn't make sense.

If the form is just:

PLR = PLR [ (EIR / EIR) + (EIR / EIR)^2, then this would mean that the efficiency has to be constant (divide by PLR on both sides).

What am I missing here?


2010/10/13 Erik Kolderup <erik at kolderupconsulting.com<mailto:erik at kolderupconsulting.com>>
The DOE2 EIR-FPLR curves include PLR in the value. In other words, you might expect the value of the curve to be (EIR at current load)/(EIR at full load), but it’s actually PLR*(EIR at current load)/(EIR at full load). So if you have a piece of equipment that has constant efficiency across the full load range, then the curve will be a straight line between (0,0) and (1,1). You have to look pretty closely at the background docs, such as the DOE2.1A Engineer’s Manual, to see that.

-Erik

Erik Kolderup, PE, LEED AP
Kolderup Consulting | 415.531.5198 | erik at kolderupconsulting.com<mailto:erik at kolderupconsulting.com>



From: James Hansen [mailto:JHANSEN at ghtltd.com<mailto:JHANSEN at ghtltd.com>]
Sent: Wednesday, October 13, 2010 1:50 PM
To: Clifford McDonald; equest-users at lists.onebuilding.org<mailto:equest-users at lists.onebuilding.org>
Subject: Re: [Equest-users] Default Performance Curves

Someone will be able to explain this better than me, but eQuest curves are not intuitive.  I wish they were, but they aren’t.

As an example:  when you enter a chiller EIR curve as a function of PLR, you would think this would be really easy, right?  0.1, 0.2, 0.3 etc as the PLR, and then the corresponding EIR (KW consumed vs current tonnage).  But that’s not the case….you end up basically having to square the EIR to get proper outputs.  Perhaps the same thing needs to be done here?

Again, someone will chime in with the real reasoning for this, as I’ve seen it discussed before.

GHT Limited
James Hansen, PE, LEED AP
Senior Associate
1010 N. Glebe Rd, Suite 200
Arlington, VA  22201-4749
703-338-5754 (Cell)
703-243-1200 (Office)
703-276-1376 (Fax)
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From: equest-users-bounces at lists.onebuilding.org<mailto:equest-users-bounces at lists.onebuilding.org> [mailto:equest-users-bounces at lists.onebuilding.org<mailto:equest-users-bounces at lists.onebuilding.org>] On Behalf Of Clifford McDonald
Sent: Wednesday, October 13, 2010 4:29 PM
To: equest-users at lists.onebuilding.org<mailto:equest-users at lists.onebuilding.org>
Subject: [Equest-users] Default Performance Curves

Hello,

I'm fairly new to eQuest, but I was looking at the default performance curves for VSDs on fans and pumps (EIR as a function of PLR), and with the exception of the cooling tower fan (htrej-Fan-Pwr-fSpeed) which is a cubic equation, they are all quadratic equations.  Why would this be when, in theory, energy use should vary with the cube of motor speed with a VSD?  Why would it be quadratic for supply fans and chilled water pumps but cubic for cooling tower fans?  Is there anywhere I can go to get more information on these default values?

Thanks for the help...
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