New Buck Corporation 261 - Manual (Page 22)

SECTION III
RESIDENTIAL FREESTANDING
REAR EXIT INSTALLATION
For optional rear exit installation locations refer back to
table of contents Section III
MINIMUM CLEARANCES TO FLOOR AND COMBUSTIBLES
See minimum floor protector measurements ,and also for minimum clearances to combustibles,
(Pages, and Figures below)
A. Rear Exit Into Masonry Flue Using Single Wall Pipe( Page23, Figure17 )
B. B. Rear Exit Into Masonry Flue Using DVL Close Clearance Pipe( Page25, Figure19 )
C. Rear Exit Vertical to Horizontal DVL Close
Clearance Pipe and Elbows
.
( Page27, Figure21 )
D. Rear Exit Into Masonry Fireplace using Single Wall Pipe
.
( Page28, Figure23,and Page29,
Figure24 )
Floor Protection:
When installing freestanding heater ,a floor protector must be use. Floor protection must be
3/8” minimum thickness non-combustible material or equivalent .R=0.06
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
protector, 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:
Thermal conductance = C = Btu = W
(hr)(ft²)(°F) (m²)(°K)
Thermal conductance = K = (Btu)(inch) = W = (Btu)
(hr)(ft²)(°f) (m)(°K) (hr)(tf)(°F)
Thermal conductance = R = (ft²)(hr)(°F) = (m²)(°K)
Btu W
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