Regency I2400M installation manual Installation, Masonry And Factory Built Fireplace Clearances

INSTALLATION

MASONRY AND FACTORY BUILT FIREPLACE CLEARANCES

The minimum required clearances to combustible materials when installed into a masonry or factory built fireplace are listed below.

	Adjacent	Mantle**	Top	Side	Minimum	Minimum	Minimum
Unit	Side Wall	(to Top)	Facing	Facing	Hearth	Hearth	Hearth Side
	(to Side)		(to Top)	(to Side)	Extension*	Thickness*	Extension
	A	B	C	D	E	F	G
Medium Flush
Insert (I2400)	11"/280mm	20"/508mm	12"/305mm	8"/200mm	18"/455mm	0.5"/13mm	8"/200mm

Side and Top facing is a maximum of 1.5" thick. Floor protection must be non-combustible, insula- tive material with an R value of 1.1 or greater.

*Hearth thickness of 0.5" with k value = 0.84", r value = 0.6 or greater.

**Anon-combustible mantel may be installed at a lower height if the framing is made of metal studs covered with a non-combustible board.

**Max. mantle depth is 10" (254mm).

Thermal floor protection is not required if the unit is raised 3.5" minimum (measured from the bottom of the stove). However, standard ember floor protec- tion is required. It will need to be a non-combustible material that covers 16" (406 mm) in the US and 18" (450 mm) in Canada to the front of the unit and 8" (200 mm) to the sides.

If the unit is not raised, thermal floor protection required is 18" (450 mm) in the US and Canada.

All floor protection must be non-combustible (i.e., metals, brick, stone, mineral fiber boards, etc.) Any organic materials (i.e. plastics, wood paper prod- ucts, etc.) are combustible and must not be used. The floor protection specified includes some form of thermal designation such as R-value (thermal resistance) or k-factor (thermal conductivity).

Clearance diagram for Installations

Minimum Hearth Extension for the front

(E)and sides (G) are measured from the fuel door opening.

HOW TO DETERMINE IF ALTERNATE

FLOOR PROTECTION MATERIALS ARE ACCEPTABLE

The specified floor protector should be 3/8" (18mm) thick material with a K - factor of 0.84.

The proposed alternative is 4" (100mm) brick with a C-factor of 1.25 over 1/8" (3mm) 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. = 1/0.29 x 0.125 = 0.431 Total R = Rbrick + Rmineral board = 0.8 + 0.431 = 1.231.

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 )(oF)			(m2))(K)
Thermal Conductivity:
k =	(Btu)(inch)		=	W	=	Btu
(hr)(ft3)(oF)			(m)(K)		(hr)(ft)(oF)
Thermal Resistance:
R	= (ft2)(hr)(oF)			=	(m2)(K)
		Btu				W

8 Regency I2400M