Floor Protection:
Floor protection must be 3/8” minimum thickness
How to use alternate materials and how to calculate equivalent thickness.
An easy means of determining if a proposed alternate floor protector meets requirements listed in the appliance manual is to follow this procedure:
1. Convert specification to
K– factor is given with a required thickness (T) in inches:
2. Determine the
Use the formula in step (1) to convert values not expressed as “R”
For multiple layers, add
3.If the overall
Example:
The specified floor protector should be 3/4” thick material with a
The proposed alternate is 4” brick with a
Step (a): Use formula above to convert specification to
Step (b): 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. =1/029 x0.125 = 0.431
Step (c): Compare proposed system R of 1.231 to specified R of 0.893. Since proposed system R is greater than required , the system is acceptable.
Definitions: |
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Thermal conductance | = C = | Btu | = | W |
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| (hr)(ft²)(°F) | (m²)(°K) |
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Thermal conductance | = K = (Btu)(inch) = | W = | (Btu) | ||
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| (hr)(ft²)(°f) |
| (m)(°K) | (hr)(tf)(°F) |
Thermal conductance = R = (ft²)(hr)(°F) = (m²)(°K)
Btu W
Install in accordance with 24 CFR, Part 3280 (HUD).
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