To make sure I am following this correctly - you need to know how long it will take to drop from 18 psi to 0 psi if you open a 2.5" hose valve on a 1902 gallon standpipe? Then - if it takes more than 3 minutes, you have to add a second hose valve?
What if you just build it and test it. Then if it takes more than 3 minutes you can cut in a mechanical T and second hose valve? It seems like there are a lot of variables that would make it impractical to do a very accurate calc. If all you need to do is add a second hose valve, why not just build it and see what happens? Matt From: Sprinklerforum [mailto:[email protected]] On Behalf Of Mike Stossel Sent: Tuesday, February 28, 2017 10:40 AM To: [email protected] Subject: RE: Air Relief Time The maximum minutes would be three the minimum can be anything, and the ambient air I would say place at about 30 since we are headed into Spring. Yes, you can definitely contact me off Forum, I greatly appreciate the help. Mike Stossel SET [cid:[email protected]] 36 Barren Road East Stroudsburg, PA 18302 Office: 973-670-2627 [email protected]<mailto:[email protected]> From: Sprinklerforum [mailto:[email protected]] On Behalf Of Brad Casterline Sent: Tuesday, February 28, 2017 11:08 AM To: [email protected]<mailto:[email protected]> Subject: RE: Air Relief Time Mike, 1) Need to know the min. and max. ambient air temperatures. 2) May I contact you off Forum? Brad ________________________________ From: Sprinklerforum [mailto:[email protected]] On Behalf Of Mike Stossel Sent: Tuesday, February 28, 2017 7:17 AM To: [email protected]<mailto:[email protected]> Subject: Air Relief Time I am trying to calculate the time it will take for a 2-1/2" hose valve to relieve a standpipe system of air. I am designing a temporary standpipe system in NYC and the requirement is to fill the entire system with air and a single 2-1/2" hose valve needs to relieve the pressure within 3 minutes or a second will need to be added. My total system volume will be 1902 gallons and the staring air pressure will be at 18psi. Does anyone know of a way to calculate this? Thanks for the help. Mike Stossel SET [cid:[email protected]] 36 Barren Road East Stroudsburg, PA 18302 Office: 973-670-2627 [email protected]<mailto:[email protected]>
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