First, forget the 4" pipe (I know there is an exception to this) but (unless the exception is met) the *MINIMUM* size of the air opening is 100 square inches. A 4" duct only has about 12 sq inches.

Let me see if I understand what you are saying:

1) There is a vent communicating the house space to the closet for combustion air, that vent is within 6" of the floor on the hall side and in the top of the 24" high platform in the closet side.

If I have that right, then the opening in the closet side is above 12" above the floor and does not meet the requirements for one of the two openings.

Also, regarding that opening, (and as Billy said) "14 x 6 = 84" so no need to go any further - it also is smaller than 100 sq inches.

2) There is a 4" duct leading up into the attic, which, as stated above, has less than 100 sq inches - which does not meet the 100 sq inches.

Okay, let's check on the opening requirements:

From the 2006 IRC (underlining is mine)

- G2407.5.3 (304.5.3) Indoor opening size and location.

Outdoor combustion air shall be provided through opening(s) to the outdoors in accordance with Section G2407.6.1 or G2407.6.2. The minimum dimension of air openings shall be not less than 3 inches (76 mm).

- - G2407.6.1 ( 304.6.1) Two-permanent-openings method.

Two permanent openings, one commencing within 12 inches (305 mm) of the top and one commencing within 12 inches (305 mm) of the bottom of the enclosure, shall be provided. The openings shall communicate directly, or by ducts, with the outdoors or spaces that freely communicate with the outdoors.

- - - Where directly communicating with the outdoors, or where communicating with the outdoors through vertical ducts, each opening shall have a minimum free area of 1 square inch per 4,000 Btu/h (550 mm2/kW) of total input rating of all appliances in the enclosure [see Figures G2407.6.1(1) and G2407.6.1(2)].
- - - Where communicating with the outdoors through horizontal ducts, each opening shall have a minimum free area of not less than 1 square inch per 2,000 Btu/h (1,100 mm2/kW) of total input rating of all appliances in the enclosure [see Figure G2407.6.1(3)].
- - G2407.6.2 (304.6.2) One-permanent-opening method. (left this off as this is not applicable to this installation)

Let's do some math:

Stated facts: "A gas burning water heater and furnace with combined BTU's of 110,000 located in a confined space "

G2407.6 says "each opening shall have a minimum free area of 1 square inch per 4,000 Btu/h (550 mm2/kW) of total input rating of all appliances in the enclosure", 110,000 / 4,000 = 27.5 X 2 = 55 square inches.

- 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.

G2407.5.3.1 says "The ratio of interior spaces shall be the available volume of all communicating spaces divided by the required volume", and the required volume is from M1702.1 Required volume: . "greater than 50 cubic feet per 1,000 Btu/h", or 110,000 / 1,000 = 110 x 50 = 5,500 cubic feet.

Now, assuming a typical house's 'openly communicating spaces' (no doors) may be 50% of the living area (just making up a number out of thin air here), and assuming a typical 1,500 sq ft house, leaving 750 sq ft 'openly communicating' (using my 50% guess) times 8 foot high ceilings = 6,000 cubic feet of 'openly communicating spaces', or, 6000 / 5500 = 1.1 is the ratio in G2407.5.3.1 above.

G2407.5.3.2 says "outdoor size reduction factor shall be one minus the ratio of interior spaces", or, 1 - 1.1 = -0.1.

G2407.5.3.3 says "minimum size of outdoor opening(s) shall be the full size of outdoor opening(s) calculated in accordance with Section G2407.6, multiplied by the reduction factor", or, 55 square inches X 0.1 reduction factor, or, 55 square inches X 0.9 = 49.5 square inches for the opening to the outdoor air (or maybe it is half that as I calculated for 2 openings - I'm getting lost reading it ).

Okay, presuming that it is only half of the above (for each opening, and there is one opening to the outdoors) then there needs to be an opening to the outdoors of 25 square inches, and the 4" round pipe gives 12 square inches ... notice that by reducing the required sized by half - it is still twice that which is what is provided by the 4" pipe (a 6" diameter pipe would have done it). Presuming *no grille or louver* on that pipe - a metal louver or grille will reduce it to 75% (give or take).

*IF* I went through that correctly (and I doubt I did), then the interior opening is wrong (too small and too high) AND the outdoor opening is too small and too low.

ONE of those openings needs to be high and ONE needs to be low - neither is 'high', the 4" pipe is low.

__________________
Jerry Peck, Construction / Litigation Consultant
Construction Litigation Consultants, LLC ( www.ConstructionLitigationConsultants.com )
www.AskCodeMan.com