England's Stove Works TRANQUILITY 17VL - Manual (Page 16)

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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:
i R‐value is given – no conversion is needed
ii k‐factor is given with a required thickness (T) in inches: R = 1/k x T
iii C‐factor is given: R = 1/C
2) Determine the R‐value of the proposed alternate floor protector:
i Use the correct formula given in step 1 (above) to convert values not expressed as
“R.”
ii For 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 = R
brick
+ R
mineral 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)(ft
2
)(deg F) (m
2
)(deg K)
Thermal conductivity = k = __(Btu)(inch)__ = ___W____ = ____Btu____
(hr)(ft
2
)(deg F) (m)(deg K) (hr)(ft)(deg F)
Thermal resistance = R = (ft
2
)(hr)(deg F) = (m
2
)(deg K)
Btu W