[Equest-users] SUPPLY-KW/FLOW

[Vikram Sami]

Yes, but I think it refers to kW/cfm

 

Vikram Sami, LEED AP 
Direct Phone 404-253-1466 | Direct Fax 404-253-1366 

LORD, AECK & SARGENT ARCHITECTURE



 

From: Brett Fero [mailto:bfero at robsonwoese.com] 
Sent: Thursday, September 17, 2009 3:08 PM
To: Vikram Sami; Lars Fetzek; equest-users at lists.onebuilding.org
Subject: RE: [Equest-users] SUPPLY-KW/FLOW

 

How about this:

 

9.9 BHP (output)/0.93 eff = 10.645 hp (input)

10.645 hp input x 0.7457 kW/hp = 7.94 kW

 

Brett Fero, P.E., LEED AP

RobsonWoese Inc.

Salina Meadows Office Park

301 Plainfield Road, Suite 180

Syracuse, NY 13212

Phone:   315-445-2650

Fax:  315-445-0958

bfero at robsonwoese.com

 

From: equest-users-bounces at lists.onebuilding.org
[mailto:equest-users-bounces at lists.onebuilding.org] On Behalf Of Vikram
Sami
Sent: Thursday, September 17, 2009 2:40 PM
To: Lars Fetzek; equest-users at lists.onebuilding.org
Subject: Re: [Equest-users] SUPPLY-KW/FLOW

 

Maybe I'm mistaken here, but I've always understood the Supply flow to
be kW/cfm. So if your design condition is 9.9 BHP, what is the design
flow rate (in cfm) that goes with that?

 

So for instance if your design flow rate is 20,000 cfm at 9.90 BHP, then
your Supply KW/Flow = 12.83/20,000 = 0.000642

 

 

 

 

Vikram Sami, LEED AP 
Direct Phone 404-253-1466 | Direct Fax 404-253-1366 

LORD, AECK & SARGENT ARCHITECTURE

 

From: equest-users-bounces at lists.onebuilding.org
[mailto:equest-users-bounces at lists.onebuilding.org] On Behalf Of Lars
Fetzek
Sent: Wednesday, September 16, 2009 4:48 PM
To: equest-users at lists.onebuilding.org
Subject: [Equest-users] SUPPLY-KW/FLOW

 

eQuesters & DOE2-2ers,

 

    The DOE2-2 dictionary loves to use the phrase, "total fan power per
unit flow" in defining SUPPLY-KW/FLOW.  Clear as mud.

    I figured this:  At design conditions, the supply fan runs at 9.90
BHP and 61.89% efficiency and the per-unit flow is at unity ( x CFM / x
CFM = 1).  The motor nominal efficiency is 93%.  (No, I don't have a
motor efficiency curve.)  0.746 kW  = 1 HP.  Therefore, the fan kW  =
9.90 * 0.746 / ( 0.6189 * 0.93 )  =  12.83 kW.  The per unit flow is 1.
So... SUPPLY-KW/FLOW  = 12.83, which I type into my .inp file.

    Naturally, eQuest gives me an error message, insisting that
SUPPLY-KW/FLOW must be between 0.0 and .0080.  Evidently, there is some
discrepancy involving units or the meaning of "per unit" in the context
of the quoted phrase.

    Can anyone clarify what eQuest really means and how to calculate
SUPPLY-KW/FLOW?  Is it supposed to be KW-per-CFM?

    Thanks!

 

Lars Fetzek, EI

Phoenix Engineering Group

Tampa, Florida


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