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APPENDIX D
3-1-2010 Page D-9
ATOMIZER PENDENT LAYOUT EXAMPLES
The follow pendent examples portray five different scenar-
ios demonstrating the application of pendent constraints.
ATOMIZER PENDENT LAYOUT EXAMPLE ONE
Example:
The size of this example area is 33 ft (10.06 m)
wide x 35 ft (10.67 m) long x 12 ft 8.5 in. (3.87 m) high and
is detailed in Figure 12.
FIGURE 12
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STEP NO. 1 – DETERMINE THE FLOOR AREA AND
GROSS VOLUME OF THE ROOM
The floor area is 33 ft (10.06 m) x 35 ft (10.67 m) which
equals 1,155 ft
2
(107.36 m
2
). The volume is calculated by
multiplying 1,155 ft
2
(107.36 m
2
) by the hazard height of
12.71 ft (3.88 m), which equals 14,680 ft
3
(416 m
3
).
This hazard does not exceed the maximum hazard volume
of 36,730 ft
3
(1040 m
3
).
STEP NO. 2 – DETERMINE THE MINIMUM NUMBER OF
ATOMIZERS REQUIRED
In this case, the ceiling height is less than 12.71 ft (3.9 m);
therefore, the number of atomizers is based strictly on the
floor area divided by the maximum rated area per atomizer
and the maximum spacing requirements.
The floor area is 1155 ft
2
(107.36 m
2
) divided by the
maximum rated area per atomizer of 289 ft
2
(26.8 m
2
)
which equals 3.99 atomizers. Round up to the nearest
whole number of 4 atomizers.
STEP NO. 3 – VERIFY ATOMIZER ARRAY PLACEMENT
Sketch the locations and verify spacing requirements of all
atomizers using the appropriate array. Additional atomizers
may be required if the spacing limitations are exceeded.
In this example it appears that a 2 x 2 rectangular array is
the most appropriate arrangement.
Typical uniform spacing of a rectangular array produces an
atomizer spacing (L) that is determined by the dimension of
the room divided by the number of atomizers in that direc-
tion and a wall spacing that is 1/2 of L.
In this case, the width of the room is 33 ft (10.06 m) divided
by two atomizers in that direction, which equals 16.5 ft (5.03
m). The spacing from the wall in this direction will be 16.5 ft
(5.03 m) divided by two, which equals 8.25 ft (2.52 m).
The length of the room is 35 ft (10.67 m) divided by 2 atom-
izers in that direction equals 17.5 ft (5.34 m). The spacing
from the wall in this direction will be 17.5 ft (5.34 m) divided
by 2 equals 8.75 ft (2.67 m).
Floor Plan Of Hazard Example Showing Rectangular
Array
FIGURE 13
008554
Examining this floor plan, the atomizer spacing require-
ments cannot be met using a 2 x 2 array because the
spacing limitations are exceeded due to the length of the
room. Because of this condition, an additional atomizer
must be added and a triangular array must be utilized.
Typical uniform spacing using a triangular array is deter-
mined by dividing the floor area of the room by the number
of atomizers which equals the floor area per atomizer.
Finally, calculate the square root of the area per atomizer to
get the distance from the center atomizer (L).
Typical Examples
35 FT
(10.67 m)
33 FT
(10.06 m)
12 FT 9 IN.
(3.87 m)
35 FT
(10.67 m)
33 FT
(10.06 m)
8 FT 3 IN.
(2.52 m)
8 FT 3 IN.
(2.52 m)
16 FT 6 IN.
(5.03 m)
17 FT 6 IN.
(5.34 m)
8 FT 9 IN.
(2.67 m)
8 FT 9 IN.
(2.67 m)