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APPENDIX D
3-1-2010
Page D-13
STEP NO. 3 – VERIFY ATOMIZER ARRAY PLACEMENT
(Continued)
In this example, first determine the spacing of the basic
pattern of five atomizers. To do this, take the floor area
divided by 5: 1225 ft
2
(113.86 m
2
) divided by 5 = 245 ft
2
(22.77 m
2
). Next, take the square root to determine (L),
which equals 15.65 ft (4.77 m). Because this is a square
room, the atomizers will form a pattern that is 22 ft (6.71 m)
on each side. This will result in a wall spacing of [35 ft
(10.67 m) – 22 ft (6.71 m)] divided by 2 = 6.5 ft (1.98 m).
The two additional atomizers will be placed on opposite
sides of the room between the corner atomizers. This will
result in a minimum atomizer spacing of 11 ft (3.35 m)
between those atomizers and the adjacent atomizers. See
Figure 19.
FIGURE 19
008560
This arrangement still allows for the repositioning of atomiz-
ers to clear obstructions.
ATOMIZER PENDENT LAYOUT EXAMPLE FOUR
Example:
The size of this example area is 32 ft (9.76 m)
wide x 52 ft (15.85 m) long x 15 ft 5 in (4.70 m) high and is
detailed in Figure 20.
FIGURE 20
008561
STEP NO. 1 – DETERMINE THE FLOOR AREA AND
GROSS VOLUME OF THE ROOM
The floor area is 32 ft (9.76 m) x 52 ft (15.85 m) which
equals 1,664 ft
2
(154.67 m
2
). The volume is calculated by
multiplying 1,664 ft
2
(154.67 m
2
) by the hazard height of
15.42 ft (4.70 m) which equals 25,659 ft
3
(726.6 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
The number of atomizers required depends on the floor
area of the room divided by the coverage area allowed for
each atomizer. To determine the rated coverage area based
on ceiling height, divide the rated volume of each atomizer
by the ceiling height
In this case the rated atomizer volume is 3673 ft
3
(104 m
3
)
divided by the ceiling height of 15.42 ft (4.70 m), which
equals an atomizer rated coverage area of 238.20 ft
2
(22.14
m
2
) per atomizer.
The number of atomizers required is 1,664 ft
2
(154.67 m
2
)
divided by 238.20 ft
2
(22.14 m
2
) which equals 6.99 atomiz-
ers. Rounding up to the nearest whole number of 7 atomiz-
ers.
Typical Examples
35 FT
(10.67 m)
35 FT
(10.67 m)
11 FT
(3.35 m)
11 FT
(3.35 m)
22 FT
(6.71 m)
22 FT
(6.71 m)
6 FT 6 IN.
(1.98 m)
6 FT 6 IN.
(1.98 m)
6 FT 6 IN.
(1.98 m)
15 FT 7 IN.
(4.77 m)
6 FT 6 IN.
(1.98 m)
35 FT
(10.67 m)
35 FT
(10.67 m)
21 FT
(6.40 m)
Summary of Contents for ANSUL AQUASONIC
Page 84: ...SECTION V 3 1 2010 Page 5 14 NOTES Installation ...
Page 88: ...SECTION VII 3 1 2010 Page 7 2 Inspection NOTES ...
Page 94: ...SECTION VIII 3 1 2010 Page 8 6 NOTES Maintenance ...
Page 116: ...APPENDIX C 3 1 2010 Page C 2 Main and Reserve Systems NOTES ...
Page 134: ...APPENDIX D 3 1 2010 Page D 18 008567 Typical Examples ...
Page 135: ...APPENDIX D 3 1 2010 Page D 19 008568 Typical Examples ...
Page 136: ...APPENDIX D 3 1 2010 Page D 20 008569 Typical Examples ...
Page 137: ...APPENDIX D 3 1 2010 Page D 21 008570 Typical Examples ...
Page 138: ...APPENDIX D 3 1 2010 Page D 22 008571 Typical Examples ...
Page 139: ...APPENDIX D 3 1 2010 Page D 23 008572 Typical Examples ...
Page 140: ...APPENDIX D 3 1 2010 Page D 24 008573 Typical Examples ...
Page 141: ...APPENDIX D 3 1 2010 Page D 25 008574 Typical Examples ...