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5
Unusually Tight Construction
The air that leaks around doors and windows may provide
enough fresh air for combustionand ventilation. However, in
buildings of unusually tight construction, you must provide
additional fresh air.
Unusually tight construction is
as con
-
struction where:
a. walls and ceilings exposed to the outside
atmosphere have a continuous water vapor
retarder with a rating of one perm (6 x 10
-11
kg
per pa-sec-m
2
) or less with openings gasketed
or sealed and
b. weather stripping 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
rs, between wall-
ceiling joints, between wall panels, at penetra
-
tions 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 Ventila -
tion Air From Outdoors, page 5.
If your home does not meet all of the three criteria
above, proceed to Determining Fresh-Air Flow for
Appliance Location, page 6.
VENTILATION AIR
Ventilation Air From Inside Building
Ventilation Air From Outdoors
Provide extra fresh air by using ventilation grills or ducts.
You must provide two permanent openings: one within
12" of the ceiling and one within 12" of the
Connect
these items directly to the outdoors or spaces open to the
.
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Follow the National Fuel Gas Code ANSI Z223.1/NFPA
54, Section 5.3, Air for Combustion and Ventilation for
required size of ventilation grills or ducts.
IMPORTANT:
Do not provide openings for inlet or outlet
air into attic if attic has a thermostat-controlled power vent.
Heated air entering the attic will activate the power vent.
Figure 1 - Ventilation Air from Inside Building
Outlet
Air
Ventilated
Attic
Outlet
Air
Inlet
Air
Inlet Air
Ventilated
Crawl Space
To
Crawl
Space
To Attic
Figure 2 - Ventilation Air from Outdoors
DETERMINING FRESH-AIR FLOW
FOR APPLIANCE LOCATION
Determining if You Ha
Use this work sheet to determine if you have a
Space: Includes the room in which you will install appli-
ance 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 = cu. ft. (volume of space)
Example: Space size 22 ft. (length) x 18 ft. (width) x
8 ft. (ceiling height) = 3168 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. Multiply the space volume by 20 to determine the maxi-
mum Btu/Hr the space can support.
_______ (volume of space) x 20 = (Maximum Btu/Hr
the space can support)
Example: 3168 cu. ft. (volume of space) x 20 = 63,360
(maximum Btu/Hr the space can support)
Содержание IVFMV18LP
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Страница 13: ...Log Placement CVO 18 24 30 MV 18 24 VFM And OVM21 13 Figure 9 Log Placement CVO MV and OVM ...
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