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]<mailto:[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
fax: 866-430-6107
email:[email protected]<mailto:email:[email protected]>

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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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