How many sprinklers do you need a room 30'x15' in ordinary hazard occupancy?

How many sprinklers do you need a room 30'x15' in ordinary hazard occupancy?

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Ok here is what I calculated. 15x30 = 450sq divided by 130sq for ordinary hazard spacing = 3.46 or rounded to 4? This was my answer I used. There was only 1 2 3 4 for the choices.

Hello update on this. I spoke with the city inspector and he said said 4 since it's an ordinary hazard. He said if it was a light hazzard it would be 2.

He is wrong, ask a sprinkler designer for an opinion, If you do please post the answer

OHHH I am not an engineer, so you can discount my reply if you want

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Cannot give a good reply, but the 130 is the maximum allowed, and than that depends on a lot of other factors.

That is why I think it is a spacing question.

OHHH I am not an engineer, so you can discount my reply if you want

A spacing question? The question asked “ how many “sprinklers. And it does not take an engineer to figure that one out. The answer is 4. If it were a “spacing” question, then the answer would be “anyway you want”. As long as you do not exceed 130 ft.² per sprinkler. And “a lot of other factors “was already decided when the question stated in “Ordinary”.

ohh, I am not an engineer and I don’t care. I’m still correct.

HI all:

If you are not a professional engineer, this chart from Sprinkler Head Spacing and Location - archtoolbox.com may help.

According to my calculation 30 ft x 15 ft = 450 sf. Based on the square foot per head in the table for ordinary hazards (130 sf), the number of heads is 450 sf / 130 sf per head = 3.46 heads. Therefore, 4 heads will cover the area for ordinary hazards.

I agree with the city's plan examiner.

Correct me if I am wrong.

I can not speculate what the local building codes and regulations are for sprinkler systems

My best,

Milt Werner

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Hi all:

To be technically correct with the standard, this is the way the standard requires the spacing determination for area/sprinkler. The sketch posted for the two heads would have S = 15 ft and L = 15 ft. S x L would be 225 sf which exceeds the limit of 130 sf for ordinary hazard. As an example for 4 heads you can use 15 ft between the branch lines for S and 8 ft between the sprinklers along the branch line for L. S x L would be 120 sf which does not exceed the limit of 130 sf for ordinary hazard. Depending on how the branch and mains are oriented on the plans, you want to keep the number (using 15 ft spacing) and lengths (shortest dimension of the area sprinklered) for branch lines to a minimize the amount of piping. Two branch lines run could across the 15 ft dimension of the room if this was the only area sprinklered.

My best,

Milt Werner

I am not an engineer

But my answer is two

If draw a 15x30 box

Put the first sprinkler in the middle of the 15 foot side 7’6” out from the long wall. So you cover 7’6”” on each side, covering the 15 foot wall

Than the next sprinkler 15 feet down from there and and once again in the middle of 15 feet

So you have the 15 side covered.

You have 22’6”” of the 30 foot covered.

Leaves only 7’6” left which is covered by the second sprinkler installed.

Draw it and check the figures.

This is using a standard coverage sprinkler

OHHH I am not an engineer, so you can discount my reply if you want