Traditions T150C, Инструкция по установке и эксплуатации

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PLANNING YOUR INSTALLATION
PAGE 5
FLOOR PROTECTION
This appliance requires ½" (13 mm) minimum non-
combustible floor protection designed for solid fuel burn-
ing appliances having a thermal conductivity of k = .84
BTU in/ft or equivalent. If the floor protection is to be
stone, tile, brick, etc., it must be mortared or grouted to
form a continuous non-combustible surface (See
Using
Alternate Material As Floor Protector below). If a chimney
connector extends horizontally over the floor, protection
must cover the floor under the connector and at least 2"
(51 mm) to either side.
The floor protection must extend completely beneath the
stove and to the front, sides, and rear as indicated:
USA REQUIREMENTS
16” min. to the front of the fuel door glass
8” min. beyond the sides of the fuel door opening
6” min. to the back of the stove body
CANADA REQUIREMENTS
18” (457 mm) min. to the front of the fuel door glass
8” (203 mm) min. beyond the sides of the stove body
8” (203 mm) min. to the back of the stove body
STANDARD PARALLEL HEARTH PAD SHOWN
USING ALTERNATE MATERIAL AS FLOOR PROTECTOR
The alternate material used as a floor protector must be
constructed of a durable noncombustible material having an
equal or better insulating value (lower k value) of k = .84
BTU / IN FT
2
HR
°
F or a thermal resistance that equals or
exceeds r = 1.19 HR
°
F FT
2
IN/BTU. With these values, de-
termine the minimum thickness/material required using the
formula and the table shown here (see chart -
Alternative
Floor Protection Materials ).
Note: Any noncombustible material having a thickness of
1/2” (12.7 mm) whose k value is less than .84 or whose r
value is more than 1.19 is acceptable. If the alternate
material used has a higher k value or lower r value will
require a greater thickness of the material used. In some
cases, if the k value is less or the r value higher, a thinner
material may be used.
Methods of determining floor protection equivalents:
To determine the thickness required for any material
when either the k or r values are known:
T
M
= Thickness of material in inches
K
M
= K value of desired material
T
L
= Minimum listed thickness
r
M
= r value of desired material
Example: Micore CV230 is to be used for the floor pro-
tection. How thick must this material be?
The following formulas give the means of determining
minimum thickness required of alternate materials.
Using the
k formula :
Desired Thickness k value of desire Minimum thickness
of the alternate = material (per inch) x of Listed
material k value of listed Material
material (per inch)
T
M
(inches) = K
M
x T
L
.84
T
M
(inches) = 0.43* x 1/2”
.84
Answer using k: 0.50 x 0.50” = 0.25 = 1/4”
1/4” thickness Micore will be required.
Using the
r formula :
T
M
(inches) = 1.19 x T
L
r
M
T
M
(inches) = 1.19 x .5”
2.33*
Answer using r: 0.50 x 0.50” = .25 = 1/4”
3/8” thickness Micore will be required.
At times it is important to know what combination of ma-
terials are acceptable for use as floor protection. The “R
values” are used to determine acceptable combinations
of materials because “R values” are additive where r and
k values are not.
“R value” = 1 = r x thickness of material used
k
ALTERNATIVE FLOOR PROTECTION MATERIALS
Values Min. Thick
k (per inch) r (per inch) T
L
Listed Material .84 1.19 1/2”
Alternative Values Min. Thick
Materials k (per inch) r (per inch) T
L
Wonderboard 1.92 0.56 1 1/8"
Common brick 5.00 0.20 3”
Cement mortar 5.00 0.20 3”
Ceramic tile 12.5 0.08 7 1/2”
Marble 11.0 0.09 6 1/2”
Micore CV230 0.43 2.33 1/4”
Ceraform 126 0.27 3.70 3/16”
Example: Given that the required “R value” for a suitable
floor protector used must be equal to or greater than:“R”
= r x T
L
= 1.19 x .5” = .60.
Note : To convert inches to millimeters divide by .03937.
8” (203 mm)
Canada
8” (203 mm)
USA
6” - USA
8” – Can.
8” (203 mm)
Canada
8” (203 mm)
USA
18” (457 mm) – Canada
16” (406 mm) - USA
f i r e - p a r t s . c o m