What is a "Projected Surface" in NFPA 15?

9/24/2024

For a flammable liquid pump surface area calculation method, NFPA 15 Section 7.3.2 says to calculate a projected prism with 5 sides or a projected plane at the footprint of the pump.

Paragraph states "projected surface" but it is not defined in the code.

What is the projected surface? How should we calculate or determine the projected surface?

Thanks in advance.

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6 Comments

Peter Howard

9/24/2024 06:45:58 am

7.3.2. Pumps, Compressors, and related Equipment.

Pumps or other devices that handle flammable liquids or gases shall have the shafts, seals, and other critical parts enveloped by directed water spray at a net rate of not less than 0.50 gpm/ sq. ft. of projected surface area of the equipment.

--

My interpretation that could very well be wrong: Anywhere the pump/compressor/or related equipment (shafts, seals, an other critical parts) is not flush with the floor or the walls, it needs to be protected at a .5 density when protected by water spray fixed systems.

I could be wrong.

Dan Wilder

9/24/2024 08:18:39 am

If you look a little further into 7.4.3.8.1, it better defines the "Projected Surface" being:

"projected horizontal or vertical plane area containing the..."

and again in 7.4.4.3.1:

"projected area of rectangular prism envelope for the..."

So my interpretation is to enclose the unit with a 5 sided box/cube (or prism) and use that surface area * .5GPM for the calculation of the required water. Any large deviations/projections may require additional boxes/prisms and the accompanying flows added for shadow protection. In it's most basic form, take the overall length/width/height of the unit being protected, multiply the sum of those 5 faces by the density, then provide that as a minimum GPM for discharge (assuming the nozzles are already discharging that as their effective axial coverage area * denisty).

In my more basic explanation of thought, think ROBLOX and the cubes created to enclose what would normally be a less flat and more rounded surface (head & body).

Jack G

9/24/2024 10:56:13 am

Agree with Dan. Must enclose surfaces when projecting the spray .

Jesse

9/24/2024 11:51:38 am

Dan's spot on here.

Franck

9/24/2024 12:51:32 pm

Dan is right

This is the same approach as for transformers (you have nice drawings in NFPA 15 with transformers, that may explain better).

A deluge protection is an object surface protection (compared to sprinklers that is a floor area protection). This means that all surfaces of the object you protect need to be wetted from the water spray discharge from the deluge nozzles. This is usually 5 sides (as the bottom side is on the ground, but could be different for a suspended tank).
This also means that the floor surface may be rather limited, but the actual area of application for your deluge system may be huge.
Imagine you need to protect an object that is a cube of 10 ft side. The floor area (that would need to be protected by sprinklers) is only 10 x 10 = 100 sq ft. But the projected surface of the object (5 sides) will be 5 times more (5 x 10 x 10 = 500 sq ft).

Glenn Berger

9/24/2024 03:33:40 pm

The example using a transformer is perfect. You base the design as if you are protecting all 5 visible sides of a box.

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What is a "Projected Surface" in NFPA 15?

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