4" with C=.72 would work Steve, thanks.
I thought the big one was 5 1/4"!

I should try to get out more.

Brad
.
On Feb 24, 2017 10:03 AM, "Steve Leyton" <[email protected]> wrote:

> Uhhhhhh … maybe it’s a 4” port?
>
>
>
> *From:* Sprinklerforum [mailto:[email protected].
> org] *On Behalf Of *Brad Casterline
> *Sent:* Friday, February 24, 2017 4:19 AM
> *To:* [email protected]
> *Subject:* Re: Do Hydrant Butts have a K Factor?
>
>
>
> Just one more thing if the Forum will suffer me--
> A while back I was looking at some flow test data and I thought I had the
> size and type of butt figured out to be 2.5" C=.9.
>
> Static= 112 psi
> Residual= 98 psi
> Pitot= 40 psi
> 4320 gpm @ 20 psi
>
> Something was bothering me because if it was a 2.5" C=.9 I get 2912 gpm @
> 20 psi.
>
> I Googled 'California hydrant butts' thinking they might have some weird
> size butts and after I got past a hot chick exercising outside didn't see
> anything unusual.
>
> So it's still a puzzle for me and I don't want to bother the poster of
> that data, hoping someone here has a simple explanation.
>
> Thanks,
>
> Brad
> .
>
> On Feb 23, 2017 3:09 PM, "Brad Casterline" <[email protected]> wrote:
>
> Good advice on using the NFPA formulas Travis, thanks, and when I am 'on
> the clock' I do.
> Thanks John. I didn't care during my first twenty years. But I always had
> an uneasy feeling when "everything I know about hydraulics I learned from
> the rules in NFPA 13". Unlike you, I was an English Major when I dropped
> out of college. Now this question is merely an after-hour curiosity, for
> entertainment purposes only if you will.
> That FM 2 1/2" K looks like C=.8 Allan, thanks.
>
> Brad
>
> On Feb 22, 2017 11:42 PM, "IPA" <[email protected]> wrote:
>
> I believe I saved this info from this very forum but I have noted that FM
> global used to have k-factors for hydrants. Here are the values:
>
> 2 3/8"
>
> 134
>
> 2 1/2"
>
> 149
>
> 2 5/8"
>
> 164
>
>
>
> On Wed, Feb 22, 2017 at 6:48 PM, John Denhardt <
> [email protected]> wrote:
>
> Travis - well said
>
>
>
> Brad - who really cares if hydrants have a K factor?  I know I don't.
>
> John August Denhardt, P.E.
>
> Strickland Fire Protection
>
>
> On Feb 22, 2017, at 8:45 PM, Travis Mack, SET <[email protected]> wrote:
>
> Or just use the formula in NFPA 291- Standard for Hydrant Marking and flow
> tests or something like that...
>
> Q = 29.83*d^2*Cd*P^0.5
>
> d = hydrant outlet diameter in inches
>
> C = the outlet coefficient (0.7, 0.8 or 0.9)
>
> P = Pitot pressure that is measured
>
> Not everything has to be extremely complicated and thought out to the nth
> degree.  If you are really so inclined, why not spend the next few years
> solving for Pi to the 1,000,000th point.  And guess what...... if you
> assign K to be 29.83*d^2*Cd........ you get 168 for your example.
> Wow...sometimes the simplest methods are the best.
>
> There is a true phenomenon that causes many people problems - *paralysis
> through analysis*.
>
>
>
> Travis Mack, SET
>
> MFP Design, LLC
>
> 2508 E Lodgepole Drive
>
> Gilbert, AZ 85298
>
> 480-505-9271 <(480)%20505-9271>
>
> fax: 866-430-6107 <(866)%20430-6107>
>
> email:[email protected]
>
>
>
> http://www.mfpdesign.com
>
> https://www.facebook.com/pages/MFP-Design-LLC/92218417692
>
> Send large files to us via: https://www.hightail.com/u/MFPDesign
>
> LinkedIn: https://www.linkedin.com/in/travismack
>
> On 2/22/2017 5:58 PM, Brad Casterline wrote:
>
> Say it's a 2.5" C=.9.
>
> I've tried to work this out a couple different ways:
> 1) a kind of long drawn out way, call it 'brad's way', and I get ~168.
> 2) a way i call the 'Bill Brooks back of napkin' way based on a few words
> he posted here a few years ago-- lightening fast, and i'll be damned if i
> dont get the same ~168! Basically, since K is the square feet area of the
> orifice he says 5*5*5.6=140, times (.9/.75) = 168.
> Anyway, if a hydrant butt has a K we could pretty quickly get the Q after
> the pitot'd P with Q=K*SQRT(P).
>
> thanks,
>
> Brad
> .
>
>
>
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