User's Manual

Design
September 2013 3-44 P/N 81-CO2MAN-001
3-6.3 Rate by Volume Method
The Rate-by-Volume application of carbon dioxide is used when the hazard is an irregular,
three-dimensional object that cannot be easily reduced to equivalent surface areas (Rate-by-
Area) and/or enclosed volumes (Total Flooding). Examples of hazards that can be protected by
this method include: printing presses, metal grinders, wave solder machines, pumps, and
motors.
The Rate-by-Volume method is applied by assuming an enclosure about the hazard. Nozzles
are located in and around the hazard to evenly distribute the agent throughout the protection
envelope and to direct the discharge at the expected flame locations. Since an enclosure does
not actually exist, the duration of protection is only as long as the agent discharge. Reference
Paragraph 3-6.1.3 for guidance on the duration of discharge.
3-6.3.1 ASSUMED ENCLOSURE
The total carbon dioxide discharge rate of the system shall be based on the volume of an
assumed enclosure, entirely surrounding the hazard. The walls and ceiling of the assumed
enclosure shall be a minimum of 2 ft. (0.6 m) from the hazard's actual edges, unless walls or
ceilings are present. The assumed enclosure shall include a solid floor.
No deductions are made to the volume for solid objects within the assumed enclosure. A
minimum dimension of 4 feet must be used during volumetric calculation of the assumed
enclosure. The assumed volume shall be increased to compensate for the losses due to winds
and forced drafts.
3-6.3.2 DISCHARGE RATE
The discharge rate for a basic system shall be 1 lb./min-
ft.
3
(16 kg/min-m
3
). If the assumed
enclosure has a closed floor and is partly defined by continuous walls that extend at least 2 ft.
(0.6 m) above the hazard and that are not part of the hazard, the minimum design flow rate
may be proportionately reduced to not less than 0.25 lb./min-
ft.
3
(4 kg/min-m
3
).
(Equation 29)
(US Units)
or
(Metric Units)
From Equation (16):
Where is the Minimum Quantity of Agent to Be Supplied and is the Duration of
Liquid Discharge.
From Paragraph 3-6.1.3:
= 0.5 min
= 1.4 x q x
= 1.4 x 89 x 0.5
= 63 lb.
W
min
1.4 qt
liq
=
W
min
t
liq
t
liq
W
min
t
liq
W
min
W
min
q
V
0.75 A
O
A
W
 0.25+=
q
V
12 A
O
A
W
 4+=