Tyco Fire Suppression & Building Products ANSUL AQUASONIC, Руководство по проектированию, установке, заправке и обслуживанию

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APPENDIX D
3-1-2010 Page D-10
STEP NO. 3 – VERIFY ATOMIZER ARRAY PLACEMENT
(Continued)
In this case the floor area is 1,155 ft
2
(107.36 m
2
) divided
by five atomizers, which equals 231 ft
2
(21.46 m
2
) per
atomizer. The square root of 231 equals 15.2 ft (4.63 m);this
is the distance (L) from the center atomizer to each of the
corner atomizers.
The location of the corner atomizers is more easily deter-
mined graphically than mathematically. In a rectangular
space, draw diagonals across opposite corners. From the
center of the room where the diagonals cross, draw a circle
of radius L (the distance from the center atomizers to the
corner atomizers). The points where the circle and the diag-
onals cross are the locations for the atomizers.
FIGURE 14
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Mathematical verification of the atomizer placement is
achieved using the following. Using the properties of similar
triangles is the simplest way to calculate the wall spacing.
The ratio of the similar triangles is determined by the
distance from the center atomizer to the corner atomizers,
divided by the distance from the center of the room to the
corners of the room. The distance from the center of the
room to the corner is the square root of [(width/2) squared
plus the (length/2) squared].
In this case, the distance from the center of the room to the
corners is (33 ft/2)
2
+ (35 ft/2)
2
= 578.5 ft
2
{(10.06 m/2)
2
+
(10.67 m/2)
2
= 53.76 m
2
}, then take the square root which
equals 24 ft (7.33 m), and the distance from the center
atomizer to the corner atomizers is 15.2 ft (4.63 m) which
was calculated previously. The ratio for the similar triangles
is therefore 15.2 ft/24 ft = 0.63 (4.63 m/7.33 m = 0.63)
Using the ratio of 0.63, the distance from the center of the
room to the atomizer location in the direction of the width is
width/2 x 0.63 = 10.4 ft (3.17 m), and the distance from the
center of the room to the atomizer location in the direction
of the length is length/2 x 0.63 = 11 ft (3.35 m). Using these
values, the wall spacing for the width is 33 ft/2 – 10.4 ft
(10.06 m/2 – 3.17 m) = 6.1 ft (1.86 m), and the wall spacing
for the length is 35 ft/2 – 11 ft (10.67 m/2 – 3.35 m) = 6.5 ft
(1.98 m).
ATOMIZER PENDENT LAYOUT EXAMPLE TWO
Example:
The size of this example area is 32 ft (9.76 m)
wide x 38 ft (11.59 m) long x 18 ft (5.49 m) high and is
detailed in Figure 15.
FIGURE 15
008556
STEP NO. 1 – DETERMINE THE FLOOR AREA AND
GROSS VOLUME OF THE ROOM
The floor area is 32 ft (9.76 m) x 38 ft (11.59 m) which
equals 1,216 ft
2
(113.03 m
2
). The volume is calculated by
multiplying 1,216 ft
2
(113.03 m
2
) by the hazard height of 18
ft (5.49 m), which equals 21,888 ft
3
(619.8 m
3
).
This hazard does not exceed the maximum hazard volume
of 36,730 ft
3
(1040 m
3
).
Typical Examples
35 FT
(10.67 m)
33 FT
(10.06 m)
6 FT 1 IN.
(1.9 m)
6 FT 1 IN.
(1.9 m)
20 FT 10 IN.
(6.4 m)
22 FT
(6.7 m)
6 FT 6 IN.
(2.0 m)
L
6 FT 6 IN.
(2.0 m)
38 FT
(11.59 m)
32 FT
(9.76 m)
18 FT
(5.49 m)