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
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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.
Mike Hill
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