Installation manual

8
Regency I2400M
INSTALLATION
Clearance diagram for Installations
Side and Top facing is a maximum of 1.5" thick.
Floor protection must be non-combustible, insula-
tive material with an R value of 1.1 or greater.
 

 



 
Thermal oor protection is not required if the unit is
raised 3.5" minimum (measured from the bottom of
the stove). However, standard ember oor protec-
tion is required. It will need to be a non-combustible
material that covers 16" (406 mm) in the US and
18" (450 mm) in Canada to the front of the unit and
8" (200 mm) to the sides.
If the unit is not raised, thermal oor protection
required is 18" (450 mm) in the US and Canada.
All oor protection must be non-combustible (i.e.,
metals, brick, stone, mineral ber boards, etc.) Any
organic materials(i.e. plastics, wood paper prod-
ucts, etc.)are combustible and must not be used.
The oor protection specied includes some form
of thermal designation such as R-value (thermal
resistance) or k-factor (thermal conductivity).

The minimum required clearances to combustible materials when installed into a masonry or factory built replace are listed below.
        
       
        
        

 11"/280mm 20"/508mm 12"/305mm 8"/200mm 18"/455mm 0.5"/13mm 8"/200mm


The specied oor protector should be 3/8"
(18mm) thick material with a K - factor of
0.84.
The proposed alternative is 4" (100mm) brick
with a C-factor of 1.25 over 1/8" (3mm) mineral
board with a K-factor of 0.29.

Use formula above to convert specication
to R-value.
R = 1/k x T = 1/0.84 x .75 = 0.893.

Calculate R of proposed system.
4" brick of C = 1.25, therefore
Rbrick = 1/C = 1/1.25 = 0.80
1/8" mineral board of k = 0.29, therefore
Rmin.bd. = 1/0.29 x 0.125 = 0.431
Total R = Rbrick + Rmineral board =
0.8 + 0.431 = 1.231.

Compare proposed system R of 1.231 to
specied R of 0.893. Since proposed system
R is greater than required, the system is
acceptable.


C = Btu = W
(hr)(ft
2
)(
o
F) (m
2)
)(K)

k = (Btu)(inch) = W = Btu
(hr)(ft
3
)(
o
F) (m)(K) (hr)(ft)(
o
F)

R = (ft
2
)(hr)(
o
F) = (m
2
)(K)
Btu W


