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
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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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