# Default Performance Curves

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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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Joined: 2011-09-30
Reputation: 0

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.

James Hansen, PE, LEED AP

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Joined: 2011-09-30
Reputation: 200

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
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 Kolderup, PE, LEED AP

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Joined: 2011-09-30
Reputation: 0

In eQUEST, I have a chilled water VAV system with no heating and an
economizer.

When I set the economizer limits to 70-F high and 55-F low, the energy
use is as expected. The chiller operates more hours, etc.

When I eliminate the low limit, the chiller energy use goes way down as
expected, however, an extremely large auxiliary energy use shows up on
the BEPS and BEPU report. It is not the pumps, because I checked the
energy use of the pumps in the PS-C report.

When looking through the output report I can't find anything that is
showing where that energy use is coming from.

Has anyone ever encountered this before that could provide some
enlightenment?

Thanks,

Dana Troy

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Joined: 2011-09-30
Reputation: 0

The physics that says energy use of a fan or pump is proportional to the
cube of the speed is for a fixed system. Most building systems are not
fixed, since there are dampers or valves that regulate flow to satisfy a
setpoint, and the speed is varied to maintain a static pressure (air
systems) or differential pressure (fluid systems). A system with perfect
pressure setpoint reset--which is all but impossible for a system serving
multiple zones--can approach the cube "law." I'm not up-to-speed on current
eQuest defaults. But for most systems, earlier default DOE-2 curves were
not conservative, but were optimistic at lower flows, partly because of the
fixed pressure the system is serving, and also because of the VFD/motor
efficiencies start to drop at low flows. There has been a number of ASHRAE
papers on system correlations. One study you may wish to review is
http://www.energy.ca.gov/2003publications/CEC-500-2003-082/CEC-500-2003-082-
A-21.PDF

William E. Koran, P.E.

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Joined: 2011-09-30
Reputation: 0

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

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Joined: 2011-09-30
Reputation: 0

I can not say for sure, but it could have to do with the way your chilled water
loop is controlled. If it is set to standby, the chiller will idle when it is
not in use. That , with a bad performance curve could possibly give you this
problem. I would try setting the loop to Demand, or put it on
a temperature schedule. Again, this is just a guess, if you post your model, I
can take a look at it.

-Eddie

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Joined: 2011-09-30
Reputation: 0

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.

[cid:image002.gif at 01CB6B8B.ABB4CC50]

John T. Forester, P.E., LEED AP

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Reputation: 0

Clifford,

When DOE2 calculates fan power, the equation is (kW current hour) = (kW at

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

it's a 16 MB file), see page IV.208.

-Erik

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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

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