Janus FM-200 Sv Series - 3.2 Agent Requirement

Page: 39
FM-200® OPERATION, DESIGN, & SERVICE MANUAL
Revision F
Document # DOC102
Issued: February 1, 2009
Revised: 11-Nov-2011
40YE
Section 3 System Design
Table 3.1.4 - Altitude Correction Chart
Altitude
Enclosure
Pressure
Correction
Factor
ft m psia mm Hg
-3,000 -914 16.25 840 1.11
-2,000 -610 15.71 812 1.07
-1,000 -305 15.23 787 1.04
0 0 14.71 760 1.00
1,000 305 14.18 733 0.96
2,000 610 13.64 705 0.93
3,000 914 13.12 678 0.89
4,000 1219 12.58 650 0.86
5,000 1524 12.04 622 0.82
6,000 1829 11.53 596 0.78
7,000 2134 11.03 570 0.75
8,000 2438 10.64 550 0.72
9,000 2743 10.22 528 0.69
10,000 3048 9.77 505 0.66
Note: Multiply the design quantity at sea level by the
correction factor to obtain the adjusted quantity for
a given altitude.
3.1.4 Hazard Altitude
FM-200® expands to a greater specic
vapor at elevations above sea level. Higher
altitudes require less agent to achieve design
concentration. Altitude dierences can be
corrected for using the correction factors
listed in Table 3.1.4.
3.2 Agent Requirement
Once the requirements and dimensions of the
hazard are determined, they can be used to
calculate the required amount of FM-200®
agent. FM-200® quantities are classied
according to storage weight. ere are two
methods to calculate the required weight.
Either the volume of the protected area can be
multiplied by an agent factor listed in Tables
3.2a or 3.2b (See Appendix B for factors and
design worksheet) or the following formula
can be used:
US Standard
W = Agent weight in pounds
V = Hazard volume in cubic feet
C = FM-200® design concentration, percent by volume
s = FM-200® specic vapor in cubic feet/pounds
s = 1.885 + (0.0046 x t)
t = minimum room temperature in °F
Example A: Our room has a volume of 16,250 ft
3
, our ambient temperature is 70°F, and our design
concentration is 6.7% (UL). Using the rst method, we consult Table 3.2a and nd our agent factor is
0.0325. Now we multiply our volume by this factor to determine the agent weight.
16,250 ft
3
X 0.0325 lbs/ft
3
= 528.13 lbs
Example B: Using the second method for the same situation, we would use the formula as follows:
s = 1.885 + (0.0046 x 70) = 2.207
16250
—
2.207
W
=
6.7
10
0
————
-
( )
———
6.7
——
=
528.74 lbs
Agent weights are always rounded up to the nearest whole pound for lling. For both Example A and
Example B this would be 529 lbs.
V
—
s
W
=
C
——
100
——
C
-
( )