User's Manual

Design
P/N 81-CO2MAN-001 3-23 September 2013
From Equation (13):
Where is the Minimum Flow Rate to Achieve 30% Within 2 Minutes.
= 0.0214 x
= 0.0214 x 4,000
= 85.6 lb./min From Equation (14):
Where is the Discharge Time.
= 
= 500 85.6
= 5.8 min
Since < 7 min, =
= 86 lb./min
EXAMPLE 8 - TOTAL FLOODING FOR DEEP-SEATED FIRES - Discharge Rate #2
Consider a dust collector having a volume of 3,000
ft.
3
. Determine the discharge rate required to
achieve a 30% concentration within 2 minutes and to complete agent discharge within 7 minutes.
From Equation (3):
Where is the Quantity of Agent for the Design Concentration, is the Quantity of Agent
to Compensate for Leakage, is the Quantity of Agent to Compensate for Ventilation, and
is the Quantity of Agent to Compensate for Extreme Temperatures.
From Equation (12):
Where is the Quantity of Agent for the Design Concentration, is the Volume of the
Enclosure, and is the Volume Factor.
= 3,000
ft.
3
= 6 ft.
3
/lb. from Table 3-3 for Dust Collectors
= 3000 6
= 500 lb.
q
30
0.0214 V=
q
30
q
30
V
q
30
q
30
t
d
W
min
q
30
=
t
d
t
d
Wmin
q
30
t
d
t
d
t
d
q
min
q
30
q
min
W
min
W
C
W
V
W
L
W
T
+++=
W
C
W
L
W
V
W
T
W
C
Vf
1
=
W
C
V
f
1
V
f
1
W
C
W
C