Vince,
I checked the friction data Ron suggested and it hit me that your original post was asking for a K Value, not K Factor! The closest thing I found was a 2.5" globe valve with a K Value of 6.1 for use in the formula hf=k(v^2/2g). Since friction loss is proportional to the square of the velocity we cannot use an equivalent feet for all flows. At 500 GPM I get k --what? = 130, and C from the previous post = .7. Instead of reducing the area, we'll reduce h, and also include the velocity head loss: 500/7.48/60=1.11 ft^3/sec, / .034 ft^2=32.77 ft/sec. hf=(32.77^2/64.4)*6.1=101.7 ft (*.433=~44 PSI). (flowing 250 GPM the Pf is 11 PSI). Velocity head, hv=(32.77^2/64.4)=16.68 ft. 100 PSI/.433=231 feet, -(101.7+16.68)=112.62 ft. v=(2*32.2*112.62)^.5=85.16 ft/sec, *.034 ft^2=2.9 ft^3/sec, *7.48*60= Q=1,300 GPM, P=100 PSI, K=130. "Theoretical": v=(2*32.2*231)^2=122 ft/sec*.034 ft^2=4.15 ft^3/sec*7.48*60=1,862 GPM. "C" = 1,300/1,862=.7 (for a 2.5 to 3" globe valve flowing 500 GPM anyway, if my thinking is correct). If anyone knows of an empirical K Value for a 2.5" hose valve please post it. A PSI/GPM curve would work, we'll just run the numbers backwards and come up with a K Value --what? -----Original Message----- From: Brad Casterline [mailto:[email protected]] Sent: Saturday, April 14, 2012 2:36 PM To: 'Vince Sabolik' Cc: [email protected] Subject: RE: k -- WHAT? Vince, I get K=149. Instead of Q=K*P^2, I used Q=vA, (Q)cubic feet per second equals (v)feet per second times (A) square feet. A=PI*radius^2,=PI*((2.5"/12/2)^2)=.034 ft^2. This is the "theoretical" area, and we have to reduce it proportional to the friction loss through the valve. For lack of a friction vs. velocity curve for now we'll assume the flow characteristics are similar to a 2.5" hydrant butt where C=.8. So A=.034*.8=.0272 ft^2. To find the velocity we use The Law of Falling Bodies, v=sqrt(2gh). g=32.2 ft/sec^2. To get h in feet we IMPLY a pressure and divide it by .433, so that if we have 100 PSI at the valve, h=100/.433=231 ft. Now, v=(2*32.2*231)^.5=122 ft/sec. Q=vA,=122 ft/sec*.0272 ft^2=3.32 ft^3/sec. K --what? in units useful to our gallon, minute, pound, programs becomes: Q=3.32 ft^3/sec,*7.48 gal/ft^3,*60 sec/minute=1,490 GPM. P=100 PSI PSI. K=Q/(P^.5)=149. We worked this out off Forum when you posted it, but based on the replies, I wanted to suggest this to the Forum at large, hoping it will help someone sometime, if it is not messed up. Also I just saw that Ron G posted a link to some friction loss data which might cure the .8 assumption, I'll check it out. ******************************************* Good morning -- Anyone have an approximate K value for a 2½" hose valve? Vince Sabolik, West Tech Fire Protection, Inc. 11351 Pearl Road / Strongsville, Ohio 44136 440 238-4800 Fax 440 238-4876 -------------- next part -------------- An HTML attachment was scrubbed... URL: <http://fireball.firesprinkler.org/mailman/private/sprinklerforum/attachment s/20120414/16e70c63/attachment.html> _______________________________________________ Sprinklerforum mailing list [email protected] http://fireball.firesprinkler.org/mailman/listinfo/sprinklerforum -------------- next part -------------- An HTML attachment was scrubbed... URL: <http://fireball.firesprinkler.org/mailman/private/sprinklerforum/attachments/20120416/4cfbb8ac/attachment.html> _______________________________________________ Sprinklerforum mailing list [email protected] http://fireball.firesprinkler.org/mailman/listinfo/sprinklerforum
