England's Stove Works 50TVL17, 50SVL17, 17VL manual Value Calculation

FLOOR PROTECTION

R Value Calculation

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
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