Continental Fireplaces CB36PTR - Page 19

18
EN
Simple venting configuration
(only one 90° elbow)
(H
T
) > (V
T
)
See graph to determine the required vertical rise V
T
for the required horizontal run H
T
.
150 (381)
147 (373)
REQUIRED
VERTICAL
100
(254)
RISE IN INCHES
(CENTIMETERS) V
T
57 (145)
50
(127)
8 (20.3)
0
5
10
15
20 (6.1)
2
(1.5)
(3.1)
12.5
(4.6)
19.5
(0.6)
(3.8)
(5.9)
HORIZONTAL VENT RUN PLUS OFFSET IN FEET (METERS) H
T
The shaded area within the lines represents acceptable values for H
T
and
V
T
For vent configurations requiring more than one 90° elbow, the following formulas apply:
Formula 1: H
T
< 4.2 V
T
Formula 2: H
T
+ V
T
< 24.75 feet (7.5m)
Example:
V
1
= V
T
= 6 FT (1.8m)
H
1
= 3 FT (0.9m)
H
2
= 5 FT (1.5m)
H
R
= H
1
+ H
2
= 3FT (0.9m) + 5FT (1.5m) = 8FT (2.4m)
H
O
= .03 (two 90° elbows - 90°) = .03 (180° - 90°) = 2.7FT (0.8m)
H
T
= H
R
+ H
O
= 8FT (2.4m) + 2.7FT (0.8m) = 10.7FT (3.3m)
H
T
+ V
T
= 10.7FT (3.3m) + 6FT (1.8m) = 16.7FT (5.1m)
90°
H 1
H 2
90°
V
1
Formula 1:
H
T
< 4.2 V
T
4.2 V
T
= 4.2FT (1.3m) x 6FT (1.8m) = 25.2FT (7.7m)
Formula 2:
H
T
+ V
T
< 24.75 FT (7.5m)
90°
90°
16.7 FT (5.1m) < 24.75 FT (7.5m)
H
1
Since both formulas are met, this vent configuration is acceptable.
H
2
Example:
V
= 4 FT (1.2m)
V
1
90°
V
2
1
= 1.5 FT (0.5m)
V
T
= V + V
2
= 4 FT (1.2m) + 1.5 FT (0.5m) = 5.5 FT (1.7m)
1
H
1
= 2 FT (0.6m)
H
2
= 1 FT (0.3m)
H
3
= 1 FT (0.3m)
H
4
= 1.5 FT (0.5m)
H
R
= H
1
+ H
2
+ H
3
+ H
4
= 2FT (0.6m) + 1FT (0.3m) + 1FT (0.3m) + 1.5FT (0.5m) = 5.5 FT (1.7m)
H
O
= .03 (four 90° elbows - 90°) = .03 (360° - 90°) = 8.1 FT (2.5m)
H
T
= H
R
+ H
O
= 5.5 FT (1.7m) + 8.1 FT (2.5m) = 13.6 FT (4.2m)
H
T
+ V
T
= 13.6 FT (4.2m) + 5.5 FT (1.7m) = 19.1 FT (5.8m)
Formula 1:
H
T
< 4.2 V
T
4.2 V
T
= 4.2 FT (1.3m) x 5.5 FT (1.7m) = 23.1 FT (7m)
13.6 FT (4.2m) < 23.1 FT (7m)
Formula 2:
H
T
+ V
T
< 24.75 FT (7.5m)
19.1 FT (5.8m) < 24.75 FT (7.5m)
Since both formulas are met, this vent configuration is acceptable.
W415-1679 / D / 03.09.18