Regency Wraps I2400M 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 fi replace 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 = 6 or greater.

**Anon-combustible mantel may be installed at a lower height if the framing is made of metalstudscoveredwithanon-combustible board.

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

Thermal fl oor protection is not required if the unit is raised 3.5" minimum (measured from the bottom of the stove). However, standard ember fl oor 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 fl oor protection required is 18" (450 mm) in the US and Canada.

All fl oor protection must be non-combustible (i.e., metals, brick, stone, mineral fi ber boards, etc.)Any organic materials (i.e. plastics, wood paper prod- ucts, etc.) are combustible and must not be used. The fl oor protection specifi ed 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

		DEFINITIONS
The specifi ed floor protector should be 3/8"	Step (b):	Thermal Conductance:
(18mm) thick material with a K - factor of	Calculate R of proposed system.
0.84.	4" brick of C = 1.25, therefore	C	=	Btu	=	W
	Rbrick = 1/C = 1/1.25 = 0.80
			(hr)(ft2 )(oF)			(m2))(K)
The proposed alternative is 4" (100mm) brick	1/8" mineral board of k = 0.29, therefore
with a C-factor of 1.25 over 1/8" (3mm) mineral	Rmin.bd. = 1/0.29 x 0.125 = 0.431	Thermal Conductivity:
board with a K-factor of 0.29.	Total R = Rbrick + Rmineral board =
Step (a):	0.8 + 0.431 = 1.231.	k =	(Btu)(inch)		= W	= Btu
	Step (c):	(hr)(ft3)(oF)			(m)(K)	(hr)(ft)(oF)
Use formula above to convert specifi cation
to R-value.	Compare proposed system R of 1.231 to	Thermal Resistance:
R = 1/k x T = 1/0.84 x .75 = 0.893.	specifiedRof0.893.Sinceproposedsystem
	R is greater than required, the system is	R = (ft2)(hr)(oF) =				(m2)(K)
	acceptable.
				Btu		W
8					Regency I2400M