Travis,

I do not believe you will be able to solve for the velocity at A with the
information provided. I think what we know at A is the sum of the velocity
head and the friction head, and it is not the Pf between A and B, but
between the source and A! By assuming one of these things we can get the
other. And the best part is we can do it using good ol' NFPA 13 only, I
think, I sketched it up and sent it to a few people asking them for
feedback. If you or any one else wants the pdf I would be happy to mail you
but I will be mostly out of the office until Thursday.
I just could not accept that 13 could not answer such a fundamentally
important question as this. Bernoulli's Theorem neglects friction, so for us
to use it we have to put Pf in and jiggle the thing around until the
formulas match the results of measurements. Bernoulli says pressure energy,
potential energy, and kinetic energy where 13 says Pt, Pn, and Pv. I also
wondered what other information is given and that is 'the water comes from
somewhere to A'!. For your 'unlimited water source' this could be a storage
tank with a fill rate of 750 GPM in this case (it would be my 'no mass is
gained or lost'). So with 3 points "T", "A", and "B" and assuming a diameter
and C-Factor and Pn at "B" = 0 (because it is discharging to atmosphere), we
can find the location between "T" and "B" where Pn at "A" is equal to Pv at
"B".

regards,
Brad Casterline
  


-----Original Message-----
From: Travis Allen [mailto:[email protected]] 
Sent: Friday, December 20, 2013 12:11 PM
To: [email protected]
Subject: RE: Flow Test Residual and Pitot The Same? 

I've spent way more time thinking and analyzing this than I should have.

Fundamentally, the answer should be yes. I say that because we are getting
hung up on the 20psi pitot. That's merely a measurement of the flow. Phrase
the question this way: Is it possible to have a static of 60 psi and a
residual of 20 psi when flowing 750 gpm?  Answer: Of course it is.

 

Hydt A is our Static/Residual and hydt B is Pitot: 

Brad helped get the juices flowing, but the first red flag was the calc'd
velocity at Hydt A at 77 ft/sec. The velocity at B can not be less, without
A having a smaller diameter.  If we assume a 6in underground, then the 77
ft/sec would equal 6,800 gpm. 

 

I think we need to go back to Bernoulli (yes, I had to spellcheck his name).
Not only is there Kinetic (Ev) and Potential (Ez) energy, but there is also
pressure energy (Ep). In addition, when dealing with SI units, we have to
include the gravitational conversion constant (remember the density in SI is
slugs/ft^3).

Hydt A is our Static/Residual and hydt B is Pitot: So Bernoulli's becomes:
Ev(hdytA) + Ez(hydtA) + Ep(HydtA) + elevation(hydt A) = Ev(hdytB) +
Ez(hydtB) + Ep(HydtB) + elevation(hydt B) + feet (friction loss between
A&B): V^2/2g + P/w + Z (@ hydt A) = V^2/2g + P/w +Z (@ hydt B) + h (friction
loss between A and B). 

At hydrant A P/w= 20/64.4= 0.31 ft  (neglible)

At hydrant B the pressure at the pitot is 0.  Ep = 0

Assume same elevation and no change in potential energy (unlimited water
source). 

Bernoulli's has now been reduced to: V^2/2g (@ hydt A) = V^2/2g(@ hydt B) +
h (friction loss between A and B). 

 

V at hydrant B is Q/a = 49 ft/sec

Therefore, at Hydt B: V^2/2g = 49^2/2(32.2) = 37.3 ft

The Velocity at the gauged Hydt A is unknown.  The energy balance is not
just elevation. The energy balance must take place at the same point in
time. Without knowing the size of the underground, or having gradient
pressure readings, I don't know how to calculate the velocity at Hydt A.  So
I'll work on this over Christmas and see if I can solve it.  Seems like a
good PE exam question..

 

Merry Christmas all.

 

Travis E. Allen, p.E.            

Sr. Fire Protection Engineer

AEGIS Engineering, PLLC

13024 Beverly Park Rd, Ste 202 , Mukilteo, WA  98275

O: 425.745.4700  | F: 206.801.5797

 <http://www.aegisengineering.com/> www.AEGISengineering.com

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