www.desatech.com 113183-01B6Unusually Tight ConstructionThe air that leaks around doors and windowsmay provide enough fresh air for combustion andventilation. However, in buildings of unusuallytight construction, you must provide additionalfresh air.Unusually tight construction is defined asconstruction where:a. walls and ceilings exposed to the out-side atmosphere have a continuouswater vapor retarder with a rating ofone perm (6 x 10-11 kg per pa-sec-m2) orless with openings gasketed or sealedandb. weather stripping has been added onopenable windows and doors andc. caulking or sealants are applied toareas such as joints around windowand door frames, between sole platesand floors, between wall-ceiling joints,between wall panels, at penetrationsfor plumbing, electrical and gas linesand at other openings.If your home meets all of these three cri-teria, you must provide additional freshair. See Ventilation Air From Outdoors,page 7.If your home does not meet all of the threecriteria above, proceed to DeterminingFresh-Air Flow For Heater Location.Confined and Unconfined SpaceThe National Fuel Gas Code, ANSI Z223.1/NFPA 54defines a confined space as a space whose volumeis less than 50 cubic feet per 1,000 Btu per hour(4.8 m3 per kw) of the aggregate input rating of allappliances installed in that space and an unconfinedspace as a space whose volume is not less than 50cubic feet per 1,000 Btu/Hr (4.8 m3 per kw) of theaggregate input rating of all appliances installed inthat space. Rooms communicating directly with thespace in which the appliances are installed*, throughopenings not furnished with doors, are considereda part of the unconfined space.* Adjoining rooms are communicating only ifthere are doorless passageways or ventilation grillsbetween them.DETERMINING FRESH-AIR FLOWFOR HEATER LOCATIONDetermining if You Have a Confined orUnconfined SpaceUse this work sheet to determine if you have aconfined or unconfined space.Space: Includes the room in which you will installheater plus any adjoining rooms with doorless pas-sageways or ventilation grills between the rooms.1. Determine the volume of the space (length xwidth x height).Length x Width x Height =__________cu. ft.(volume of space)Example: Space size 20 ft. (length) x 16 ft.(width) x 8 ft. (ceiling height) = 2560 cu. ft.(volume of space)If additional ventilation to adjoining room issupplied with grills or openings, add the volumeof these rooms to the total volume of the space.2. Multiply the space volume by 20 to determinethe maximum Btu/Hr the space can support.__________ (volume of space) x 20 = (Maxi-mum Btu/Hr the space can support)Example: 2560 cu. ft. (volume of space) x 20 =51,200 (maximum Btu/Hr the space can support)3. Add the Btu/Hr of all fuel burning appliances inthe space.Vent-free heater ___________ Btu/HrGas water heater* ___________ Btu/HrGas furnace ___________ Btu/HrVented gas heater ___________ Btu/HrGas fireplace logs ___________ Btu/HrGas water heater __________ Btu/HrOther gas appliances* + __________ Btu/HrTotal = __________ Btu/Hr* Do not include direct-vent gas appliances. Di-rect-vent draws combustion air from the outdoorsand vents to the outdoors.Example:Gas water heater __________ Btu/HrVent-free heater + __________ Btu/HrTotal = __________ Btu/Hr4. Compare the maximum Btu/Hr the space cansupport with the actual amount of Btu/Hr used._________ Btu/Hr (maximum the space can support)_________ Btu/Hr (actual amount of Btu/Hr used)Example: 51,200 Btu/Hr (maximum the spacecan support)79,000 Btu/Hr (actual amount ofBtu/Hr used)The space in the above example is a confined spacebecause the actual Btu/Hr used is more than the maxi-mum Btu/Hr the space can support. You must provideadditional fresh air. Your options are as follows:AIR FOR COMBUSTIONAND VENTILATIONContinued40,00039,00079,000