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MAGNUMZC MANUAL – April 2006
PAGE - -
17
HOW TO DETERMINE IF FLOOR PROTECTION MATERIALS ARE ACCEPTABLE
ü
All floor protection materials must be non-combustible (i.e., metals, brick, stone, mineral fiberboard,
etc.). Any organic materials (i.e., plastics, wood, paper products, etc.) are combustible and must not
be used.
ü
The floor protector specified may include some form of thermal designation such as R-value (thermal
resistance), or K-factor (thermal conductivity), or C-factor (thermal conductance).
3/8” thick rock
board or mineral wool board meets these specifications.
The technical means of determining if a proposed alternate floor protector meets requirements listed in the
appliance manual is to follow this procedure
:
a)
Convert specification to R-value:
•
R-value given – no conversion needed.
•
K-value is given with a required thickness
(T) in inches:
R = 1/K X T (1)
•
C-value is given:
R
=
1/C
(2)
b)
Determine the R-value of the proposed alternate floor protector.
•
Use the formula in step (a) to convert values not expressed as “R.”
•
For multiple layers, add R-values of each layer to determine overall R-value.
c)
If the overall R-value of the system is greater than the R-value of the specified floor protector, the
alternate is acceptable.
EXAMPLE:
The specified floor protector should be ¾ inch thick material with a K-value of .84. The proposed
alternate is 4” brick with a C-value of 1.25 over 1/8” mineral board with a K-value of .29.
Step (a):
Use formula (1) to convert specification to R-value. R + 1/K X T + 1/.84 X .75 + .893
Step (b):
Calculate R of proposed system.
4”
brick
of
C
=
1.25
Rbrick
=
1/C
=
1/1.25
=
.80
1/8’
mineral
board
of
K
=
.29
Rmin.bd.
=
1/.29
X
.125
-
.431
Total
R
=
Rbrick
+
Rmineral
board
=
.8
+
.431
=
1.231
Step (c):
Compare proposed system of R of 1.231 to specified R of .893. Since proposed system R
is greater than required, the system is acceptable.
Simply put, 3/8” thick rock board or mineral wool board will meet these specifications.
Thermal Conductance
= C = Btu
(hr) (ft2) (F)
Thermal Conductivity
= k = (Btu) (in)
(hr)
(ft
2
) (F)
or
Btu
(hr) (ft) (F)
Thermal Resistance
=
R = (ft2) (hr) (F)
Btu