New Buck Corporation 261 SECTION II RESIDENTIAL FREESTANDING VERTICAL EXIT INST ALLATION, Floor Protection:

SECTION II RESIDENTIAL FREESTANDING

VERTICAL EXIT INSTALLATION

MINIMUM CLEARANCES TO FLOOR AND COMBUSTIBLES

See minimum floor protector measurements ,and also for minimum clearances to combustibles,

See Pages, and Figures below

A. Vertical Exit Single Wall Pipe( Page9, Figure3 )

B. Vertical Exit DVL ( Page11, Figure5 )

C. Vertical Wall Exit ,Single Wall Pipe, Elbow,103 HT Chimney and T-BOX(Page13, Figure7)

D. Vertical Wall Exit , DVL Pipe, Elbow,103 HT Chimney and T-BOX( Page15, Figure9 )

E. Vertical Exit to Masonry Flue Single Wall Pipe and Elbow( Page17, Figure11)

F. Vertical Exit to Masonry Flue DVL Pipe and Elbow( Page19, Figure13 )

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