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 85298480-505-9271 <(480)%20505-9271>
>>> fax: 866-430-6107 <(866)%20430-6107>email:[email protected]
>>> http://www.mfpdesign.comhttps://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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