If I didn't have a playpipe, I followed FM's conservative guidelines (from
their Hydraulic Pocket Guide). 0.80 for outlets that were smooth and well
rounded inside. And 0.70 when the outlet was square.

 

Travis

 

 

From: Travis Mack, SET [mailto:[email protected]] 
Sent: Wednesday, February 22, 2017 5:45 PM
To: [email protected]
Subject: Re: Do Hydrant Butts have a K Factor?

 

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]
 
http://www.mfpdesign.com
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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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