Sure, but why assume anything? Why not calculate it to the bottom of the
tank? A flow test is: Static--- PSI/.433=h ft. pitot is velocity
pressure/.433=h ft giving the water the velocity--- times the sq ft of the
hydrant butt gives ft^3/sec-- (the .8 or .9 comes from the coefficient of
constriction-- the diameter of the stream is less than the opening)-- this
would be the diameter of the pipe at the bottom of the tank. The difference
with a flow test is it is steady pressure (therefore steady flow)-- if the
tank had steady fill rate with overflow 'drain', it would be the same thing,
because h would be constant. 

-----Original Message-----
From: Ralphy Henderson [mailto:[email protected]] 
Sent: Tuesday, June 15, 2010 2:40 PM
To: [email protected]
Subject: RE: Gravity Tanks

Couldn't you also do this:

For assumption sake let's say 1st floor is light hazard and we're
anticipating we're going to need 400 gpm. Plot out the static pressure of
the tank (at the empty level) over zero gpm then calculate our friction
losses through the main from the water tank down to the first floor based
upon our anticipated 400 gpm and subtract that from our static to get our
residual pressure. Plot out the residual pressure over 400 gpm and we now
have a standard water supply curve that can be used to determine other
pressures at available flows.

--- On Tue, 6/15/10, Brad <[email protected]> wrote:

From: Brad <[email protected]>
Subject: RE: Gravity Tanks
To: "'Matt Grise'" <[email protected]>, [email protected]
Date: Tuesday, June 15, 2010, 5:44 PM

For turbulent (sprinkler) flows, friction loss is proportional to the square
of the velocity

-----Original Message-----
From: Matt Grise [mailto:[email protected]] 
Sent: Tuesday, June 15, 2010 12:41 PM
To: '[email protected]'; '[email protected]'
Subject: RE: Gravity Tanks

Don't forget friction in the pipes!

Matt Grisé PE*, LEED AP 
Sales Engineer 
Alliance Fire Protection 
*Licensed in KS & MO 

913.888.0647 ph 
913.888.0618 f 
913.927.0222 cell 
www. AFPsprink.com 


-----Original Message-----
From: [email protected]
[mailto:[email protected]] On Behalf Of Brad
Sent: Tuesday, June 15, 2010 12:39 PM
To: [email protected]; [email protected]
Subject: RE: Gravity Tanks

i.e., if the water level is 280 feet 'above your head', then h=280 ft, v=134
ft/sec (same as the Law of Falling Bodies- if you were dropped from 280 ft,
you would be going 134 ft/sec when you hit the ground).
ft/sec*ft^2=ft^3/sec. If the flow is coming out of 4" pipe, ft^2=.09----
*134= 12 * 7.48= 90---- * 60= 5400 GPM. As the level 'above your head'
drops, so does the velocity, so either use calculus, or call h the bottom of
the tank, like I would have to do. It still seems like magic to me that by
just using feet and seconds, given the rated psi and rpm ONLY, of a fire
pump, I can calculate the diameter of the impeller (v=2*pi*r*f). It seems
like magic bcoz for 15 years I tried to learn sprinkler calcs after the PEs
had simplified it enough for me by factoring gravity out of the formulas.   

-----Original Message-----
From: Brad [mailto:[email protected]] 
Sent: Tuesday, June 15, 2010 8:30 AM
To: [email protected]
Subject: RE: Gravity Tanks

mgh=.5mv^2
m is the same on both sides so 
v= sq root 2gh
h= v^2/2g
first thing is forget about gallons, minutes, and psi- use feet and seconds:
ft, ft^2, ft^3, sec, sec^2. ft/sec=(ft^3/sec)/ft^2)
g=32.2 ft/sec^2 (if this project is on the earth). 7.48 gal/ft^3. water
weighs 62.4 lbs/ft^3.

-----Original Message-----
From: Todd Williams [mailto:[email protected]] 
Sent: Tuesday, June 15, 2010 5:11 AM
To: [email protected]
Subject: Gravity Tanks

Was there a thread on calculating flow from a gravity tanks a while 
back? I couldn't find it. I have to do a calculation on the first 
floor of a building fed from a gravity tank on the 28th floor

Todd G. Williams, PE
Fire Protection Design/Consulting
Stonington, CT
860.535.2080
www.fpdc.com

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