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-14
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.
This hazard area is fairly long and somewhat narrow.
Combined with the fact that seven atomizers are required,
this indicates that a combination of rectangular and triangu-
lar arrays may be needed. Since the room is only 32 ft
(9.76 m) wide, two atomizers can be used to satisfy the
spacing across the width. However, the room is too long for
three atomizers across the length. Use the triangular array
with five of the atomizers to extend the room length that can
be covered without adding an additional atomizer to get to a
2 x 4 array.
Calculate the area coverage for each atomizer of 1,664 ft
2
(154.67 m
2
) divided by 7 atomizers which equals 237.7 ft
2
(22.10 m
2
). Using two of the atomizers across the width on
one end of the room will cover an area of 475.4 ft
2
(44.20
m
2
). Divided this area by the width of the room of 32 ft (9.76
m), and 14.85 ft (4.53 m) of the room has been covered.
This leaves an area of 37.15 ft (11.33 m) by 32 ft (9.76 m)
that still needs coverage.
Using the same process in Example 1, take the square root
of the area per atomizer from above, 237.7 ft
2
(22.10 m
2
),
to get the atomizer spacing dimension (L) which equals
15.42 ft (4.70 m). Draw diagonals for the remaining area
and a circle of radius L at the intersection of the diagonals.
Position the atomizers at each of the intersections.
FIGURE 21
008562
Mathematical verification of the atomizer placement is
achieved using the following steps. 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 atomiz-
ers, 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 remaining
area to the corners of the remaining area is [(32 ft/2)
2
+
(37.15 ft/2)
2
] = 601 ft
2
{(9.76 m/2)
2
+ (11.33 m /2)
2
= 55.91
m
2
}, then take the square root of 24.52 ft (7.48 m), and the
distance from the center atomizer to the corner atomizers is
15.42 ft (4.70 m) which was calculated previously. The ratio
for the similar triangles is therefore 15.42 ft/24.52 ft = 0.63
(4.70 m/7.48 m = 0.63)
Using the ratio of 0.63, the distance from the center of the
remaining area to the atomizer location in the direction of
the width is: width/2 x 0.63 = 10.1 ft (3.08 m), and the
distance from the center of the remaining area to the
atomizer location in the direction of the length is: length/2 x
0.63 = 11.7 ft (3.57 m). Using these values, the wall spacing
for the width is 32 ft/2 – 10.1 ft (9.76 m/2 – 3.08 m) = 5.9 ft
(1.8 m), and the wall spacing for the length is 37.15 ft/2 –
11.7 ft (11.33 m/2 – 3.57 m)= 6.9 ft (2.1 m).
This layout can be simplified by moving the atomizers width-
wise that form the corners of the triangular array in by
approximately 1.5 ft (0.46 m), and moving the two atomizers
of the rectangular array out width-wise by 0.5 ft (0.15 m).
FIGURE 22
008563
Typical Examples
52 FT
(15.85 m)
37 FT 2 IN.
(11.33 m)
14 FT 10 IN.
(4.53 m)
32 FT
(9.76 m)
11 FT 8 IN.
(3.56 m)
11 FT 8 IN.
(3.56 m)
15 FT 5 IN.
(4.7 m)
7 FT 5 IN.
(2.26 m)
7 FT 5 IN.
(2.26 m)
6 FT 11 IN.
(2.11 m)
6 FT 11 IN.
(2.11 m)
10 FT 1 IN.
(3.07 m)
10 FT 1 IN.
(3.07 m)
5 FT 11 IN.
(1.80 m)
5 FT 11 IN.
(1.80 m)
52 FT
(15.85 m)
17 FT
(5.18 m)
23 FT 4 IN.
(7.11 m)
14 FT 4 IN.
(4.40 m)
14 FT 4 IN.
(4.40 m)
32 FT
(9.76 m)
11 FT 8 IN.
(3.56 m)
8 FT 6 IN.
(2.59 m)
7 FT 5 IN.
(2.26 m)
6 FT 11 IN.
(2.11 m)
7 FT 6 IN.
(2.26 m)
7 FT 6 IN.
(2.26 m)