Specifications; Water Vapor: A By-Product Of Unvented Room Heaters; Provisions For Adequate Combustion & Ventilation Air - White Mountain Hearth Vail Installation Instructions And Owner's Manual

VFP24FP3*L
Model
VFP24FP7*L
Input Max. 20,000
Min. 14,000
Fireplace Dimensions
Height 26 3/8"
Depth 10 7/8"
Front Width 22 5/8"
Rear Width 17 1/4"
Gas Inlet 3/8"
* Indicates optional accessories that may be installed.

WATER VAPOR: A BY-PRODUCT OF UNVENTED ROOM HEATERS

Water vapor is a by-product of gas combustion. An unvented room
heater produces approximately one ounce (30ml) of water for every
1,000 BTU's (.3KW's) of gas input per hour. .
Unvented room heaters are recommended as supplemental heat (a
room) rather than a primary heat source (an entire house). In most
supplemental heat applications, the water vapor does not create a
problem. In most applications, the water vapor enhances the low
humidity atmosphere experienced during cold weather.
PROVISIONS FOR ADEQUATE COMBUSTION & VENTILATION AIR
This heater shall not be installed in a confined space or unusu-
ally tight construction unless provisions are provided for adequate
combustion and ventilation air.
A confined space is an area with volume less than 50 cubic feet per
1,000 Btuh of the combined input rates of all appliances drawing
combustion air from that space. Small areas such as equipment
rooms are confined spaces. Furnaces installed in a confined space
which supply heated air to areas outside the space must draw return
air from outside the space through tightly sealed return air ducts. A
confined space must have 2 openings into the space for combus-
tion air. One opening must be within 12 inches of the ceiling and
the other must be within 12 inches of the floor. The required sizing
of these openings is determined by whether inside or outside air is
used to support combustion, the method by which the air is brought
to the space (vertical or horizontal duct) and by the total input rate
of all appliances in the space.
Unusually Tight Construction
The air that leaks around doors and windows may provide enough
fresh air for combustion and ventilation. However, in buildings of
unusually tight construction, you must provide additional fresh air.
Unusually tight construction is defined as construction
where:
a. Walls and ceilings exposed to the outside atmosphere have
a continuous water vapor retarder with a rating of one perm
or less with openings gasketed or sealed, and
b. Weatherstripping has been added on openable windows and
doors, and
c. Caulking or sealants are applied to areas such as joints
around window and door frames, between sole plates and
floors, between wall-ceiling joints, between wall panels, at
Page 6

SPECIFICATIONS

VFP24FP2*L VFP24FP3*L10
20,000 10,000
10,000 10,000
26 3/8" 26 3/8"
10 7/8" 10 7/8"
22 5/8" 22 5/8"
17 1/4" 17 1/4"
3/8" 3/8"
Accessories
FBB6 Automatic Blower
VPP1A-22 Aged Brick Liner
VFF24BR-1 Brass Trim Kit
VFF24SS-1 Stainless Steel Trim Kit
VFF24HP-1 Hammered Pewter Trim Kit
Accessories for Millivolt and Intermittant Models Only
FRBC Battery Operated Remote Control
FRBTC Battery Operated Remote Control w/Thermostat
FREC Electric Remote Control
FWS Wall Switch
TMV Wall Thermostat, Millivolt - Reed Switch
TRW Wireless Remote Wall Thermostat
The following steps will help insure that water vapor does not
become a problem.
1. Be sure the heater is sized properly for the application, including
ample combustion air and circulation air.
2. If high humidity is experienced, a dehumidifier may be used to
help lower the water vapor content of the air.
3. Do not use an unvented room heater as the primary heat
source.
penetrations for plumbing, electrical, and gas lines, and at
other openings.
If your home meets all of the three criteria above, you must provide
additional fresh air. See "Ventilation Air From Outdoors," page
7.
Determining if You Have a Confined or Unconfined Space
Use this worksheet to determine if you have a confined or unconfined
space.
Space: Includes the room in which you will install heater plus any
adjoining rooms with doorless passageways or ventilation grills
between the rooms.
1. Determine the volume of the space (length x width x height).
Length x Width x Height =
space)
Example: Space size 16 ft. (length) x 10 ft. (width) x 8 ft. (ceiling
height) = 1,280 cu. ft. (volume of space)
If additional ventilation to adjoining room is supplied with grills
or openings, add the volume of these rooms to the total volume
of the space.
2. Divide the space volume by 50 cubic feet to determine the
maximum BTU/Hr the space can support.
(volume of space) ÷ 50 cu. ft. = (maximum
BTU/Hr the space can support)
Example: 1,280 cu. ft. (volume of space) ÷ 50 cu. ft. = 25.6 or
25,600 (maximum BTU/Hr the space can support)
cu. ft. (volume of
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