New Buck Corporation FS 21 installation instructions Floor Protection

Floor Protection:

Floor protection must be 3/8” minimum thickness non-combustible material or equivalent.

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 R-value:

R-value is given—no conversion is needed.

K-factor is given with a required thickness (T) in inches: C-factor is given: R=1/C

2. Determine the R-value of the proposed alternate floor protector.

Use the formula in step (1) to convert values not expressed as “R”

For multiple layers, add R-values of each layer to determine the overall R-value.

3.If the overall R-value of the system is grater than the R-value of the specified floor protec- tor, the alternate is acceptable.

Example:

The specified floor protector should be 3/4” thick material with a K-factor of 0.84.

The proposed alternate is 4” brick with a C-factor of 1.25 over 1/8” mineral board with a K-factor of 0.29.

Step (a): Use formula above to convert specification to R-value. R= 1/K x T = 1/0.84 x .75 = 0.893

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:				Btu		W
Thermal conductance	=	C	=		=
				(hr)(ft²)(°F)		(m²)(°K)
Thermal conductance	=	K	=	(Btu)(inch)	=	W	=	(Btu)
				(hr)(ft²)(°F)		(m)(°K)		(hr)(ft)(°F)
Thermal conductance	= R =			(ft²)(hr)(°F)	=	(m²)(°K)
				Btu		W

Install in accordance with 24 CFR, Part 3280 (HUD).

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