Well, you won't be testing the BFP during that fire situation, either, so that 
removes a problem of that transient variable :)


George L.  Church, Jr., CET  
Rowe Sprinkler Systems, Inc.
PO Box 407, Middleburg, PA 17842
877-324-ROWE       570-837-6335 fax
[email protected]


-----Original Message-----
From: [email protected] 
[mailto:[email protected]] On Behalf Of Vince Sabolik
Sent: Monday, April 16, 2012 2:40 PM
To: [email protected]
Subject: RE: k -- WHAT?

Forum --

This was my question, so I'd like to clarify a little.

OK, yes this IS for full flow testing a backflow preventer. It was based on 
being able to calculate a performance to save some pipe size.

AND yes, I was looking for an approximate K "value" for a 2½" valve.

IF I had to guess, that would be a value of 149, plus or minus.

BUT since all hydraulics are pretty much based on a 50 degree temperature (back 
in the 1800's the folks that were solving these problems standardized on that 
temperature), how can any of this be "precise"
when you're not gonna find 50 degree water in a fire situation in the first 
place?

As far as a K Factor, K Value - whatever - goes, I was always taught that the 
value represented a sprinkler, a group of sprinklers, etc. etc. etc.  And as 
all of us know, K= GPM/SQR(P).  The formula that was developed originally used 
an expotential of 1.85
(logarithm) instead of 2, so the results are off to begin with, but close 
enough for rock n roll.

Most of us (me) can't do log, so we used square root in many, many hand calcs.

I suppose the advent of computers had impact on this, but now we're back to 
being exact - for 50 degree water in a fire area.

It may vary a little bit from manufacturer to manufacturer, but given that 
extact thing, if a K Factor can represent a head, group of heads etc., why 
couldn't  it represent a 2½" valve's performance as well?




At 01:51 PM 4/16/2012, you wrote:
>Wasn't it a hose valve for forward flushing of the BFP that was the 
>original question?
>
>And don't K factors have to be empirically derived? I once tried to 
>determine the logo sprinkler atop a building wherein I worked in 1988, 
>and was told without drilling a hole where the orifice was and pumping 
>a LOT of water thru it, we couldn't determine it via formula. I was 
>talking to folks in Lansdale that I think know a lot about large K 
>factors, even back in the 80's.
>
>If I were to try to determine how many 2.5" FHV I needed to flow a 
>certain value, I'd simulate it with a test header. Can't think its much 
>different than the # of FHVs needed per #20, and we've been doing that 
>far longer than we've been using even hand calcs.
>
>KISB. Keep It Simple, Brad. This is an imprecise business, deal with it.
>
>
>George L.  Church, Jr., CET
>Rowe Sprinkler Systems, Inc.
>PO Box 407, Middleburg, PA 17842
>877-324-ROWE       570-837-6335 fax
>[email protected]
>
>
>-----Original Message-----
>From: [email protected]
>[mailto:[email protected]] On Behalf Of Brad 
>Casterline
>Sent: Monday, April 16, 2012 11:37 AM
>To: [email protected]
>Subject: RE: k -- WHAT?
>
>But since friction head with turbulent
>(sprinkler) flow is proportional to the velocity squared, and the data 
>from Ron Greenman's link says 'K Value for use in the formula 
>hf=k(v^2/2g)', and the velocity varies as the flow, wouldn't the K 
>Value give us the friction loss 'curve' we are looking for, as opposed 
>to an average over a range?
>
>-----Original Message-----
>From: Roland Huggins [mailto:[email protected]]
>Sent: Monday, April 16, 2012 10:00 AM
>To: [email protected]
>Subject: Re: k -- WHAT?
>
>I started to say before getting overly precise (but that horse has left 
>the barn), it might be helpful to keep in mind that the K valve on a 
>sprinkler is NOT a constant at different flows.  It is an AVERAGED 
>result for flows from 7 psi to 100 psi.
>
>Roland
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