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CLEARANCE TO COMBUSTIBLES
The following stated clearances represent the minimum
distances between the stove and any other object. No
objects should encroach into this space. This includes
but is not limited to carpet, furniture, children, pets,
clothing, fuel, or any other object. These clearances
may only be reduced by means approved by the
regulatory authority having jurisdiction.
REAR WALL
D
A
G
Figure 1 Clearance to Combustibles
USA
A 9" (229 mm)
B 9.4" (239 mm)
C 15.9" (404 mm)
D 7.9" (201 mm)
E 3" (77 mm)
F 5.9" (150mm)
G 19" (483mm)
FLOORING SPACE & CLEARANCES
When installed on a combustible floor, non-
combustible floor protection is required to:
• Cover the area beneath the stove and extend at
least 15.9 inches (404 mm) to the front
• Cover the area at least 7.9 inches (201 mm) beyond
each side and 5.9 inches (150mm) beyond rear of
the room heater.
• Cover the area under the exhaust venting and 2
inches (50.8 mm) beyond each side.
Additionally, the wood pellet fire stove shall be
positioned such that:
• It has at least 9" (229 mm) of clearance from the
each side to the nearest body.
Cleveland Iron Works Wood Pellet Fire Stove
D
FRONT OF HEATER
NON-COMBUSTIBLE
FLOOR PROTECTION
CANADA
9" (229 mm)
9.4" (239 mm)
15.9" (404 mm)
7.9" (201 mm)
3" (77 mm)
5.9" (150mm)
19" (483mm)
• It has at least 9.4" (239 mm) of clearance from the
rear to the nearest body.
• Vertical runs of vent pipe must be at least 3" (77
mm) from any wall.
Finally, the area which the wood pellet fire stove is
installed shall have a floor-to-ceiling distance of at least
84" (2134 mm).
FLOORING MATERIAL
Floor protection must be all of the following:
• Listed to UL 1618.
• At least 0.5" (13 mm) thick
• Constructed of non-combustible material.
• Have either:
Thermal resistance value R of 1.19
Thermal conductivity value k of 0.84
For assistance evaluating the suitability of substitute
materials, the following equivalences of specifications
and example below have been provided.
Thermal conductivity k =
Thermal conductance C =
Example: Required to protect floor with R value of 1.19
(ft
2
)(hr)(
0
F)
Btu
.
Evaluating merit of 2¼ inch (57 mm) thick brick with
thermal conductivity k = 4.16
(6.3 mm) thick mineral board that has C value of 2.3
(Btu)
(ft
)(hr)(
F)
2
0
.
Step 1. Calculate the R value of each floor material
thickness
k
R
=
BRICK
1
C
R
=
BOARD
Step 2. Add the equivalent R values for each floor
material
R
+ R
= 0.54 + 0.434 = 0.974
BRICK
BOARD
Step 3. This combined R value is insufficient and so
more protection must be provided. For example,
by using 2 layers of bricks:
R
+ R
+ R
BRICK
BRICK
BOARD
Step 4. Because this combined R value is larger than
the specification, this is a sufficient method for
protecting the floor area underneath the stove.
5
Operating Instructions and Owner's Manual
(ft
)(hr)(
F)
2
0
Btu
(Btu) (inch)
(ft
2
)(hr)(
0
F)
(Imperial or SI units)
thickness
(Btu) (inch)
(
R
(ft
2
)(hr)(
0
F)
or
1
(Btu)
(
R
(ft
2
)(hr)(
0
F)
or
(Btu) (inch)
(ft
)(hr)(
F)
2
0
on top of ¼ inch
2.25
4.16
=
= 0.54
1
2.3
=
= 0.434
= 0.54 +0.54 + 0.434 = 1.514
W
)
(m)(
0
K)
W
)
(m
2
)(
0
K)
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