Maximum sprinkler spacing refers to "Protection area of coverage" and this is also used for determining a head's required discharge.
"Area of coverage" is the actual physical floor area covered by the head. This value is used when calculating the remote area size.
Trevor Spain Igneus Incorporated On 12/3/2017 7:41 PM, Bruce Verhei wrote:
I’d give the original poster *****’s for putting up a practical question of interest. Best. Bruce VerheiOn Dec 3, 2017, at 17:11, [email protected] wrote: I have no doubt Roland will shoot this thing in the head by Lunch Time in Kansas, Tomorrow. Richard Bradley Casterline. Quoting "MFP Design, LLC" <[email protected]>:No, you don’t ignore the obstruction. I agree with that. I was just pointing out that SxL is not always the same for spacing and calculations. Travis Mack, SET MFP Design, LLC 3356 E Vallejo Ct Gilbert, AZ 85298 480-505-9271 fax: 866-430-6107 <mailto:email:[email protected]> email:[email protected] <https://na01.safelinks.protection.outlook.com/?url=http%3A%2F%2Fwww.mfpdesign.com%2F&data=02%7C01%7C%7C1121d49f9e6b4cf248f108d4df580e77%7C14e5497c16da42e69ffa77d19bafe511%7C0%7C0%7C636379016677342180&sdata=HJ8OA4xyeHAoxXNz5mu%2FYfycgtd5nsFrrpvzulZiNkQ%3D&reserved=0> http://www.mfpdesign.com <https://na01.safelinks.protection.outlook.com/?url=https%3A%2F%2Fwww.facebook.com%2Fpages%2FMFP-Design-LLC%2F92218417692&data=02%7C01%7C%7C1121d49f9e6b4cf248f108d4df580e77%7C14e5497c16da42e69ffa77d19bafe511%7C0%7C0%7C636379016677342180&sdata=H%2BwdcgK8DLGBcNoqJEvUrzsXngySwkX56Vgf9gM9EGk%3D&reserved=0> https://www.facebook.com/pages/MFP-Design-LLC/92218417692 Send large files to us via: <https://na01.safelinks.protection.outlook.com/?url=https%3A%2F%2Fwww.hightail.com%2Fu%2FMFPDesign&data=02%7C01%7C%7C1121d49f9e6b4cf248f108d4df580e77%7C14e5497c16da42e69ffa77d19bafe511%7C0%7C0%7C636379016677342180&sdata=eGdMZGu2wXhUupGwgGTrqF3b54OP5%2BAZvlHhABSexWY%3D&reserved=0> https://www.hightail.com/u/MFPDesign LinkedIn: <https://na01.safelinks.protection.outlook.com/?url=https%3A%2F%2Fwww.linkedin.com%2Fin%2Ftravismack&data=02%7C01%7C%7C1121d49f9e6b4cf248f108d4df580e77%7C14e5497c16da42e69ffa77d19bafe511%7C0%7C0%7C636379016677342180&sdata=tT5E7LsZjSmyreKi4gDCa70EWN%2BZodi%2FhbeCbHNRijI%3D&reserved=0> https://www.linkedin.com/in/travismack “The bitterness of poor quality remains long after the sweetness of low price is forgotten.” From: Sprinklerforum [mailto:[email protected]] On Behalf Of Kyle.Montgomery Sent: Sunday, December 3, 2017 2:03 PM To: [email protected] Subject: RE: [EXTERNAL] Re: Saw-Tooth Ceilings That is true. (See 23.4.4.6.5) Thanks for pointing that out. But that does not mean you can ignore the obstruction when determining the required discharge, right? That was the premise of the original question. I’ll amend my previous statement to say the area used for determining minimum discharge should be APPROXIMATELY 210 square feet, based on the original question and the precise slope of the ceiling. -Kyle M From: Sprinklerforum [mailto:[email protected]] On Behalf Of MFP Design, LLC Sent: Sunday, December 03, 2017 12:03 PM To: [email protected] <mailto:[email protected]> Subject: RE: [EXTERNAL] Re: Saw-Tooth Ceilings Not necessarily. Remember the generic phrase: Space on the slope, calc on the flat. So, you may be at 130 sq ft on the slope, but if it translates to 110 sq ft on the flat, then you calc for 110 sq ft. From: Sprinklerforum [mailto:[email protected]] On Behalf Of Kyle.Montgomery Sent: Sunday, December 3, 2017 10:45 AM To: [email protected] <mailto:[email protected]> Subject: RE: [EXTERNAL] Re: Saw-Tooth Ceilings I disagree, Brad. Whatever distances you use to determine S x L for spacing the sprinklers, the same distances must be used to determine area of coverage for hydraulic calculations. (NOTE: There are a few exceptions, such as when using the 'Small Room Rule'). I think that NFPA 13 spells it out pretty clearly: 23.4 Hydraulic Calculation Procedures 23.4.4 Calculation Procedure. 23.4.4.6 Design Densities 23.4.4.6.3: The area covered by any sprinkler used in hydraulic design and calculations shall be the horizontal distances measured between the sprinklers on the branch line and between the branch lines IN ACCORDANCE WITH 8.5.2. 8.5.2 Protection Areas per Sprinkler: 8.5.2.1: Determination of Protection Area of Coverage. 8.5.2.1.1 The protection area of coverage per sprinkler shall be determined as follows: 1. … 2. Between branch lines as follows: a. Determine the perpendicular distance to the sprinkler on the adjacent branch line (or to a wall OR OBSTRUCTION in the case of the last branch line) on each side of the branch line on which the sprinkler is positioned. b. Choose the LARGER of either twice the distance to the wall OR OBSTRUCTION or the distance to the next sprinkler. c. Define dimension as L. I think that based on these code sections, the answer to the original question is that the sprinklers should be calculated as A = S x L = 15’ x (2 x 7’) = 210 square feet. So minimum discharge would be 21 GPM for light hazard. And if it’s to be considered Ordinary Hazard, the heads should be spaced at about 9’-3” along the branch lines. -Kyle M -----Original Message----- From: Sprinklerforum [mailto:[email protected]] On Behalf Of [email protected] <mailto:[email protected]> Sent: Thursday, November 30, 2017 3:15 PM To: [email protected] <mailto:[email protected]> Subject: [EXTERNAL] Re: Saw-Tooth Ceilings Space the sprinklers like you were plumb-bobbing up. Calculate the density like you were plumb-bobbing down. B. Quoting Michael Hill < <mailto:[email protected]> [email protected]>:So you are saying that this is a series of 15 sloped ceilings one after the other, not a single obstructed saw tooth ceiling. I am not trying to be argumentative. I am trying to convince myself that what I am being told to do is correct. Mike Hill -----Original Message----- From: Sprinklerforum [ <mailto:[email protected]> mailto:[email protected]] On Behalf Of <mailto:[email protected]> [email protected] Sent: Thursday, November 30, 2017 4:02 PM To: <mailto:[email protected]> [email protected] Subject: Re: Saw-Tooth Ceilings 23.4.4.6 Deisign Densities 23.4.4.6.2, and for backup- 23.4.4.6.5. You don't have walls Michael, you have a sloped ceiling. The down-slope 'half' of the pattern is obstructed, be it sawtooth or not :). Brad. Quoting Michael Hill < <mailto:[email protected]> [email protected]>:That section explains what to do once the maximum area of protection is determined. I am still not convinced. I agree that plenty of water is going to get to the fire, if needed. More so as this is a UFC project, HC-2 occupancy, dry system, 0.20/3500. Suppose there were walls that extended down to the floor at each end of the ceiling that created a compartment (which is how I am viewing this situation). In this case the compartment would be 240' x 8'. 8.5.2 tells me to take the distance from the wall, double it and multiply by the distance between sprinklers in the other direction. No mention of if the area is broken down into smaller compartments. Does anyone know of a section in NFPA-13 that allows me to ignore the obstruction when determining area of protection per sprinkler. This is how I look at things when I do not agree with them. I figure I am sitting on a witness stand somewhere (after the fire the destroyed the building) and some insurance company's' lawyer is saying "point to where it says you can do that". Mike Hill -----Original Message----- From: Sprinklerforum [ <mailto:[email protected]> mailto:[email protected]] On Behalf Of <mailto:[email protected]> [email protected] Sent: Thursday, November 30, 2017 2:34 PM To: <mailto:[email protected]> [email protected] Subject: Re: Saw-Tooth Ceilings I'm with Ron. See NFPA 13-2016 23.4.4.6.2. Brad. Quoting Ron Greenman < <mailto:[email protected]> [email protected]>:I'll agree the sprinkler knows nothing, but that is irrelevant. We are working with a mathematical model that will, if all the rules are followed, give us a reasonable degree of confidence that it will perform to the design intended, which is, in this case, to keep a fire contained within the most demanding design area (or any other similar, but less demanding area) for a specified duration. So for OH I can space my sprinklers along the ceiling at a MAXIMUM of 15 ft apart, in any direction perpendicular to the walls, which is a maximum square footage per sprinkler of 225 square feet, and a MINIMUM of six feet between them for a total square footage of 36. So given Mike's layout, he is well within the parameters of this rule. Now though we run into another rule which states we cannot EXCEED 130 square feet of floor area per sprinkler in OH so if we run our sprinklers at 15' apart on the line we can have a MAXIMUM distance between lines a fraction over 8.5 feet. Since the sawtooth creates a "compartment," so to speak, because of the obstructive nature of the "wall" the "compartment" distances are 8x15 per sprinkler so he is well within the spacing rules regarding "throw," and since his horizontal space coverage area at the floor is less than 130 sqft per sprinkler he is within the area limitation for OH. His density will be a minimum of 0.15 gallons/min/sqft and presuming he has enough pressure to push that quantity of water through the system to the 13 (before adjustments to the design area, if any) design sprinklers the system will p[erform as desired. Back to the ignorant sprinkler. Will it adjust to the 1.0'x7.5'x7.0'x7.5' layout and deliver exactly 0.15 gpm to each square foot per minute? Of course not. What will happen though is the system will supply enough water to up to 13 sprinklers anywhere in the system to meet the design parameter (containment of the fire to no more than 1500sqft). We can feel confident about this since sprinkler systems have been doing so for over a hundred years both in pipe schedule and hydraulically designed configurations. In every system, there is a prize and that is containment or suppression of a fire. There are several design methods to obtain the prize and 13 is merely a prescriptive set of rules to get us there. As long as we follow the rules we'll be alright. If we want to play mathematician or fluid dynamicist then we'll cease to be layout technicians and we'll get distracted and our eyes will stray from the prize. On Thu, Nov 30, 2017 at 9:33 AM, Nick Maneen < <mailto:[email protected]> [email protected]> wrote:You are correct. The sprinkler does not know it needs to throw water in mainly one direction. It will be 15x14 *Nick Maneen, SET * *c* 704.791.7789 <(704)%20791-7789> *From:* Sprinklerforum [mailto:[email protected]. org] *On Behalf Of *Michael Hill *Sent:* Thursday, November 30, 2017 12:30 PM *To:* <mailto:[email protected]> [email protected] *Subject:* RE: Saw-Tooth Ceilings Yes, the actual floor area is about 120ft² per sprinkler. Each sprinkler has to spray 7’-0” towards the next line. The next sprinkler is within 1’-0” of the vertical portion of the ceiling (which I am considering an obstruction) and will not spray any further in that direction. The water will hit the obstruction and fall straight down. I know the sprinklers can cover that 7’-0” spray, but I need to make sure it does so at the correct density. So for hydraulic purposes is the spacing considered 120ft² or is it 210ft². My take is that the sprinkler is 7’-0” form a wall or obstruction, as the sprinkler has to spray 7’-0” to cover to the low portion of the saw-toothed ceiling. L=14 and S=15. So hydraulically that sprinkler has to cover 210ft². I would like to say it is covering 120ft² but I need to see where in NFPA-13 that it says I can do so. Mike Hill *From:* Sprinklerforum [mailto:[email protected]. org < <mailto:[email protected]> [email protected]>] *On Behalf Of *Ron Greenman *Sent:* Thursday, November 30, 2017 12:09 PM *To:* <mailto:[email protected]> [email protected] *Subject:* Re: Saw-Tooth Ceilings If I'm understanding your description and question properly you are covering about 116-117 sqft of floor area if one considers the 1'8" rise at the tip of the tooth an obstruction, which I would, and your "throw distance" along the slope is at 0-360 degrees=7.5'/90 degrees=7.0'/180 degrees=7.5 degrees/270 degrees=1.0'. That puts you within coverage rules for OH, and throw range rules for SS sprinklers. On Thu, Nov 30, 2017 at 8:33 AM, Michael Hill < <mailto:[email protected]> [email protected]> wrote: I am working on a firing range that is going to have a saw-tooth type ceiling installed. I have a question about determining maximum protection area per sprinkler. The vertical portion is 1’-8” tall with the distance from the high point to the low point being 8’-0”, measured along the slope. Sprinklers need to be spaced close to the vertical portion of the ceiling, to protect them from stray projectiles. The sprinkler will end up being about 7’-0” from the low point of the ceiling, measured along the slope. The sprinklers are about 8’-0” apart, but with a 1’-8” vertical portion of the ceiling between them. Sprinklers are spaced 15’-0” apart along the branchline. See attached sketch (if it came through). Would the spacing be 15’ x 8’ or would it be 15’ x 14’? I feel it should be 15’ x 14’ as any individual sprinkler needs to cover 7’-0” in one direction, measured along the slope. I feel the vertical portion of the ceiling is an obstruction and can almost been interpreted as a wall that extends to the floor. I also thought I needed to increase my design area for sloped ceiling for this situation, but have since been shown the commentary in the handbook that explains why it is not required for saw-toothed ceilings. Maybe I am wrong about the spacing as well. 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