Originally Posted by
Paul Cobb
I'm new to the forums and my expertise is in the electrical field but I have a plumbing problem. I replaced an existing 40,000 BTU gas water heater at my house in southeast Florida. It is in an enclosure approx. 6'x3'x6' high on the outside of the building. The local inspector wants me to drill (20) 1 inch holes in the roof of the enclosure and (20) 1 inch holes in the walls of the enclosure. Can anyone explain to me why this is necessary? Doing this will turn the enclosure into an indoor water catchment area. Thanks for any input you can give me. If there is an online code reference for me to lookup that would be great.
.
Paul,
I have no idea where that inspector came up with those desires (certainly not requirements by the code). As soon as you drill those holes in the roof, you are now making the water heater basically installed "outside", and that is not listed, labeled, designed or intended for installation "outside".
This is from the Florida Building Code, Residential.
- M1702.1 Required volume.
- - Where the volume of the space in which fuel-burning appliances are installed is greater than 50 cubic feet per 1,000 Btu/h (4.83 L/W) of aggregate input rating in buildings of ordinary tightness, insofar as infiltration is concerned, normal infiltration shall be regarded as adequate to provide combustion air. Rooms communicating directly with the space in which the appliances are installed through openings not furnished with doors shall be considered part of the required volume.
- M1702.2 Confined space.
- -Where the space in which the appliance is located does not meet the criterion specified in Section M1702.1, two permanent openings to adjacent spaces shall be provided so that the combined volume of all spaces meets the criterion. One opening shall be within 12 inches (305 mm) of the top and one within 12 inches (305 mm) of the bottom of the space, as illustrated in Figure M1702.2. Each opening shall have a free area equal to a minimum of 1 square inch per 1,000 Btu/h (2.20 mm
2/W) input rating of all appliances installed within the space, but not less than 100 square inches (0.064 m
2).
6'x3'x6' = 108 cubic feet of space within the enclosure
40,000 BTU @ 50 cubic feet per 1,000 Btu/h = 50 x 40 = 2,000 cubic feet of space needed to not be considered "confined space"
That space does not meet the requirements of M1702.1 (above), thus that space is considered "confined space", so you go to M1702.2 (also above).
However, being as all air will be taken from outdoors, you go to M1703.1: (underlining is mine)
- M1703.1 Outdoor air.
- - Where the space in which fuel-burning appliances are located does not meet the criterion for indoor air specified in Section M1702, outside combustion air shall be supplied as specified in Section M1703.2.
- M1703.2
Two openings or ducts.
- - Outside combustion air shall be supplied through openings or ducts, as illustrated in Figures M1703.2(1), M1703.2(2), M1703.2(3) and M1703.2(4).
One opening shall be within 12 inches (305 mm) of the top of the enclosure, and
one within 12 inches (305 mm) of the bottom of the enclosure. Openings are permitted to connect to spaces directly communicating with the outdoors, such as ventilated crawl spaces or ventilated attic spaces. The same duct or opening shall not serve both combustion air openings. The duct serving the upper opening shall be level or extend upward from the appliance space.
- - - M1703.2.1 Size of openings.
- - - - Where communicating with the outdoors by means of vertical ducts, each opening shall have a free area of at least 1 square inch per 4,000 Btu/h (0.550 mm2/W) of total input rating of all appliances in the space.
Where horizontal ducts are used, each opening shall have a free area of at least 1 square inch per 2,000 Btu/h (0.275 mm2/W) of total input of all appliances in the space. Ducts shall be of the same minimum cross-sectional area as the required free area of the openings to which they connect.
The minimum cross-sectional dimension of rectangular air ducts shall be 3 inches (76 mm).
That means you will need 1 square inch per 2,000 Btu/h, being as the water heater is 40,000 Btu/h, that means you need 20 square inches net free vent area for each opening. Using metal louvers (you will need a louver to keep the rain out), and with metal louvers being rated at 75% of their size for net free vent area, you would need an opening with a louver which is at least 27 square inches (27 x 75% = 20.25 square inches net free vent area).
An opening with the minimum allowable dimension of 3" means the other opening dimension would need to be 9", making the opening 3" x 9" = 27 cubic inches x 75% = 20.25 square inches per opening x 2 openings, one within 12" of the top of the enclosure and one within 12" of the bottom of the enclosure. While the openings *could* start at the 12" mark, I always recommend the openings end at the 12" mark. I.e., instead of coming down 12" for the top of the top opening and making the opening 3" down from there, come down 12" for the bottom of the top opening, making the opening going up from there. The reverse applies to the bottom opening. That way, if you screw it up and 'miss your 12" mark', you do not have to make the openings larger, you are still "within" the 12" of the top and bottom.
If your local inspector has questions on the code and its requirements, have them call me.
Maybe he will teach me something I don't know (or maybe it will be the other way around
).
Go here (
Florida Building Codes Home Page ) click on Florida Building Code 2004, scroll down to 2004 Florida Building Code: Residential (2nd link down), scroll down to Chapter 17 - Combustion Air, scroll down to the section I posted above. That link also includes the figures referred to in the code.
