When we calculate 4 adjacent sprinklers in a nfpa 13 system (residential section) we're suppose to calc the four sprinks which create the most demanding condition. I believe the handbook uses the word 'contiguous' to further describe the situation but my question is this:

You have a line with six sprinklers on it. The spacing for each head (because they're in separate rooms) is the last two sprinklers are at 324 sq. ft., the two next to it are at 256 sq. ft. followed by two more at 324 sq. ft. next to the main.

Looking similar to this:

----X-----X-----X-----X-----X-----X

....324....324...256..256...324...324


Because we're talking about 'adjacent' sprinklers I would calculate the last four sprinklers on the line which would include two at 324 sq. ft. and two at 256 sq. ft. right? Or should the four sprinklers include all four of the 324 sq. ft.? I think the former is correct but just want to make sure.



Thanks,

The last sprinkler on the line will drive the pressure of the lower square footage sprinklers behind it. If the situation was reversed, then for simplicity I would calculate all at the higher square footage.

Fire Protection Engineers give Firefighters sloppy seconds.

Right Pyrophoric,

But now let's say the 256 sq. ft. sprinklers are 5.6K and the 324 sq. ft. are 8.0K - because they're in different size rooms the larger k-factors are not needed in the smaller sq. ft. areas. It seems to me you would still have a mix of 256 & 324 sq. ft. sprinklers either the last four sprinklers or the beginning four but not just the four 324 sq. ft. sprinklers because they're not technically 'adjacent'. Right?

The hydraulics should always show the most demanding condition.
Let's take a 0.10 density per sprinkler:
32.4 gpm k8 = 16.4025psi
25.6 gpm k5.6 = 20.898 psi

If the last sprinkler on a line generates the most pressure, and every other sprinkler has a larger k factor, you will end up wasting a ton of water to balancing, as you can see the k8 will get 4psi more than minimum design and discharge 36+/-gpm. You may have to calculate multiple areas to determine what the actual most-demanding area is, and if it reads "adjacent" then take the most demanding continous four heads on a line.

Fire Protection Engineers give Firefighters sloppy seconds.

Calc programs are so simple with today's software. Just run both and see which one is the most demanding. That will solve the issue.

Travis Mack
MFP Design, LLC
www.mfpdesign.com

Travis,

You don't even need a calculator to figure out which sprinklers are going to be more demanding in the scenario I'm presenting - the issue is more about the words 'adjacent' and 'contiguous'. Does that mean the sprinkler next in line on the branchline has to be included in the calculations because it's adjacent to the most demanding head - or should you just start grabbing the four most demanding sprinklers which, in the illustration above, would be the (4) 324 sq. ft. sprinklers even though that selection would probably be better defined as incongruous than contiguous?

You always take adjacent sprinklers, even in an area design. Find the most demanding adjacent 4. You would not make a 3000 square foor area full of holes.

Fire Protection Engineers give Firefighters sloppy seconds.

Just my opinion but I would think the intent is to calculate the most demanding adjacent 4 sprinklers on a common line. The point being to suppress a fire in one area along the line. I believe the word contiguous means sharing a boundary (i.e. contiguous lower 48 state of the USA) so they would be the 4 installed on the same line even if in different rooms. I might even submit a few different calculations to satisfy the AHJ that the system is acceptable.