[Equest-users] Default Performance Curves

Jonathan Smith jsmith at csarchpc.com
Fri Oct 15 06:22:14 PDT 2010


Clifford, 

I have also wondered why the default fan VSD curve is a quadratic
equation when the 90.1 baseline VSD fan curve is cubic. To compare the
quadratic and cubic formulas, I used the default eQuest equation and
coefficients for comparison to those provided in ASHRAE 90.1 Method 2
(Table G3.1.3.15). Please see the attached comparison.

Figure 37 in the DOE2.2 help file under SUPPLY-KW/FLOW may also prove
useful.


Jonathan R. Smith AIA LEED(r)AP
CSArch
Architecture|Engineering|Construction Management
40 Beaver Street
Albany, New York 12207
p: 518.463.8068 x268
f: 518.463.8069 
www.csarchpc.com



-----Original Message-----
From: Erik Kolderup [mailto:erik at kolderupconsulting.com] 
Sent: Thursday, October 14, 2010 1:14 PM
To: 'Clifford McDonald'
Cc: equest-users at lists.onebuilding.org
Subject: Re: [Equest-users] Default Performance Curves

Clifford,

 

When DOE2 calculates fan power, the equation is (kW current hour) = (kW
at
full load)*(EIR-FPLR curve at current part load ratio). 

 

So the output of the EIR-FPLR curve is defined as PLR*(EIR at current
PLR)/(EIR at full load). And PLR is (current hour airflow)/(peak design
airflow). The form of the EIR-FPLR curve can be linear, quadratic or
cubic
and the choice is up to you (the defaults are typically quadratic or
cubic).
If it's quadratic then the form is EIR-FPLR = a + b*PLR + c*PLR^2. If,
for
example, you have constant EIR across the load range, then coefficients
a
and c would be zero and b would be 1. Then you have a linear curve
EIR-FPLR
= 1*PLR. 

 

Since that's probably clear as mud, another place to look for more info
is
the engineers manual

http://doe2.com/download/DOE-21E/DOE-2EngineersManualVersion2.1A.pdf
(note
it's a 16 MB file), see page IV.208. 

 

-Erik

 

From: Clifford McDonald [mailto:mcdonald.cliff at gmail.com] 
Sent: Thursday, October 14, 2010 6:09 AM
To: Erik Kolderup
Cc: James Hansen; 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>

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

 

 

 

From: James Hansen [mailto:JHANSEN at ghtltd.com] 
Sent: Wednesday, October 13, 2010 1:50 PM
To: Clifford McDonald; 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)

 <http://www.ghtltd.com/> www.ghtltd.com

 

 

From: 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
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...

  _____  

The information contained in this communication is confidential, may be
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the
property of GHT Limited.  Unauthorized use, disclosure or copying of
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communication or any part thereof is strictly prohibited and may be
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this communication and all copies thereof, including all attachments.
Thank
you.

 


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