Thanks Skylar, all too experienced in venting from previous bad experiences 
back in the day.
The underground we did yesterday we fortunately had 2 hydrants and the 8" 
flanged stub in the pump room to vent.  The stub was where we flushed from (to 
get the entire main) and also pressure tested/forward flow tested from.  I 
welded up a set of manifolds years ago with 2.5" grooved weldolets that take 
grooved hose valves.  We vented the hydrants and also through the manifold and 
even with a slight dip in elevation between hyd. 1 and 2 before entering the 
building I think we got 'most' of the air.  I can usually tell in the first 10 
minutes, depending on volume, if we got trapped air.  I use a good calibrated 
gauge (pump gauge) and saw an initial drop from 210 to 207 in the first 10 
minutes but rock solid after to 2 hours.  

These roadway babies are all over the place but, we have AAV's at every 
elevated station.  6 coupling expansion joints.  The  AAV's have ball valves on 
them so I figure 2 passes end to end should vent, check the hose valves for 
tight an caps tight, close the ball valves on the AAV's; all under street 
pressure.  The trick, well one of them, is going to be checking for leaks, like 
I’m going to hang my fat head over the side 75' in the air, and then find a 
pump suitable for the initial charge, and maybe recharge.  Looking.  

I also have a clause in discharging dirty water that I didn't before, plus one 
for noise.  We are close to the harbor and old Ironsides so we may have to run 
hoses to a sanitary manhole, we still have catch basins that drain to the 
ocean.  I'm excited to do this, despite the conditions, I'll be taking pictures 
and video unless the Commonwealth says no.  They did during the big dig.
Any advice or help is appreciated guys.  


Scot - I don't think your base equation is quite right.  You may be trying to 
use the ideal gas law here, on a liquid, which doesn't work.  Also, I think 
technically you would be changing "n" in the equation, not V (the volume of the 
system is constant at P1 and P2), but I don't think this really matters.  I 
think your equation is essentially correct, but only if we were talking about a 
system being filled and tested with a gas, not water.

Tom - I have considered this problem in the past, when I had to test a run of 
underground piping in the middle of nowhere, and I wanted to know how big of a 
tank I would need to bring to get to 200 psi after the piping was full of 
water.  Because water does not compress very much (see link below for a graph; 
less than .5% at 1000 PSI), the answer was extremely small (less than 10 
gallons).  I don't want to run through the math again (it doesn't really matter 
because we are talking about only needing to add a very small volume of water 
to increase the pressure substantially).  The huge key is having no air in the 
system.  Air is very much compressible, as shown in Scot's equation.  I say all 
of this because my advice is that you should vent out as much air as possible.  
If you can do this, it shouldn't take much time to pressure up the system up to 
200 psi, even with a small pump.  Getting all the air out is usually easier 
said than done (theory can be much different than real life), but with what I 
just told you, it is worth making an attempt.

https://www.quora.com/How-much-force-would-be-needed-to-compress-water-within-an-indestructible-vessel-And-what-would-happen


Thanks,
Skyler Bilbo

1700 S. Raney Street
Effingham, IL 62401
217-819-6404 Cell

[email protected]
www.wenteplumbing.com


On Fri, Apr 16, 2021 at 3:27 AM å... .... via Sprinklerforum < 
[email protected]> wrote:

> To answer your question:
>
> *P1 *V1*    =  P2 * V2    when tested at close to the same temperature.
> V2  =    (P1*V1)/P2      where  P1 = 200 psi/14.7 psi = 13.6 atm,  V1 =
> volumen of your pipe system,  P2 is  1 atm.
> V2 =  13.6 * V1
> If we have a 4 gpm pump, replace V2 with  (4 gpm * x min), rearrange 
> to solve for minutes.  If we have 2 pumps operating in parallel, we 
> use the flow rate of the pump with the lower developed-pressure.  V1 is in 
> gallons.
>
> X minutes to fill  =  (13.6 * V1 ) / 4 gpm * FF
>
> FF is a fudge factor.  It accounts for leaks, the fact that your pump 
> may not be exactly 4 gpm, the fact your pump flow rate will decrease 
> as the pressure it pushes against increases, and FF  accounts for plain and 
> simple
> entropy.   My hunch is FF is about 1.3.   But don't take my word for it,
> record the results yourself and show how bad this hunch is.
>
> "The inspector came back after the requisite 2 hours and signed it off.
> It's
> all about relationships."
>

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