England's Stove Works 50-SFP12L, 50-TFP12P Installation on a Combustible Floor, Wall Protection

Installation on a Combustible Floor

If the appliance is to be installed on a combustible floor or a combustible floor covering, it must be installed on a 1” thick non-combustible millboard floor protector or a durable equivalent, with a “R” factor of no less than “2.” The pad must be installed beneath the unit, extending 16” (U.S.) on the side equipped with a door, and 8” on all other sides. The pad must cover any horizontal chimney connector runs and extend 2” beyond each side.

Alternate Floor Protection:

An easy means of determining if a proposed alternate floor protector meets requirements is to follow this procedure:

1)Convert specification to R-value:

iR-value is given – no conversion is needed

iik-factor is given with a required thickness (T) in inches: R = 1/k x T

iiiC-factor is given: R = 1/C

2)Determine the R-value of the proposed alternate floor protector:

iUse the correct formula given in step 1 (above) to convert values not expressed as “R.”

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

3)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” 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 R brick = 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

Step (c): Compare proposed system of R of 1.231 to specified R of 0.893. Since proposed system R is greater than required, the system is acceptable.

Definitions:
Thermal conductance	= C		=	Btu	=	W
				(hr)(ft2)(deg F)		(m2)(deg K)
Thermal conductivity	=	k	=	(Btu)(inch)	=	W	=	Btu
				(hr)(ft2)(deg F)		(m)(deg K)		(hr)(ft)(deg F)
Thermal resistance	=	R	= (ft2)(hr)(deg F)		=	(m2)(deg K)
				Btu		W

2. Wall Protection

Please see Illustration 1 for clearances to walls. In some areas local codes may require thirty- six inches (36”) from a combustible, therefore it is very important that you check with local officials. If you need to place your unit closer to a combustible wall, some protection will be necessary. If an approved wall board is used this will reduce your clearance by two thirds (2/3); however, a one inch (1”) air space has to be between the board and the wall. If you have a ceiling flue hook-up, you will need protection from the floor to the ceiling if you do not meet the normal clearances. If you have a wall flue hook up, you will need wall protection at least twelve inches (12”) above the wall thimble.

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