## Sample Units:

Furnace 80,000 BTU + Water Heater 40,000 BTU = 120,000 BTU Total

### Minimum Indoor Air Communicating with the Appliances:

The minimum required volume shall be 50 cubic feet per 1,000 Btu/h of the appliance input rating. ( (Total BTU Input of all Appliances / 1,000) x 50 = Volume Required )

(120,000 BTU / 1,000) x 50 ft3 = 6,000 ft3

#### Combustion Air Supplied From Same Story:

The minimum free area of 1 square inch per 1,000 Btu/h of the total input rating of all appliances in the enclosure.
(Total BTU Input of all Appliances / 1,000 = Free Area of Each Opening Required )

(120,000 BTU / 1,000) x 1 in2 = 120 in2 minimum free area for each of the required (2) openings

Utilizing metal louvers requires (120 in2/.75 = 160 in2) to obtain 120 in2 of free area.

Confirmation: 75% of 160 in2 = .75 x 160 in2 = 120 in2

One opening 160 in2 within 12 inches of the top and one opening 160 in2 within 12 inches of the bottom of the enclosure would meet the minimum requirements.

Note: the appliances must be communicating with a minimum of 6,000 ft3 of space as calculated above.

#### Combustion Air Supplied From Different Story:

The minimum free area of 2 square inches per 1,000 Btu/h of total input rating of all appliances in the enclosure.

( Total BTU Input of all Appliances / 1,000 x 2 = Total Free Area Required )

(120,000 BTU / 1,000) x 2 in2 = 240 in2 minimum free utilizing one or more openings

Utilizing metal louvers requires (240 in2/.75 = 320 in2) to obtain 240 in2 of free area.

Confirmation: 75% of 320 in2 = .75 x 320 in2 = 240 in2

One or more openings in doors or floors 320 in2 communicating with a different story would meet the minimum requirements.

### Minimum Outdoor Air Communicating with the Appliances:

#### Two-permanent-openings method:

##### Vertical Ducts:

The minimum free area of each opening shall have a minimum free area of 1 square inch per 4,000 Btu/h of total input rating of all appliances in the enclosure.
(Total BTU Input of all Appliances / 4,000 = Total Free Area Required )

(120,000 BTU / 4,000) x 1 in2 = 30 in2 minimum free area for each of the required (2) openings

Assume duct is communicating with the attic space and louvers are not utilized.

One opening 30 in2 within 12 inches of the top and one opening 30 in2 within 12 inches of the bottom of the enclosure would meet the minimum requirements.

##### Horizontal Ducts:

The minimum free area of each opening shall have a minimum free area of 1 square inch per 2,000 Btu/h of total input rating of all appliances in the enclosure.
( Total BTU Input of all Appliances / 2,000 = Total Free Area Required )

(120,000 BTU / 2,000) x 1 in2 = 60 in2 minimum free area for each of the required (2) openings

Utilizing metal louvers requires (60 in2/.75 = 80 in2) to obtain 60 in2 of free area.

Confirmation: 75% of 80 in2 = .75 x 80 in2 = 60 in2

One opening 80 in2 within 12 inches of the top and one opening 80 in2 within 12 inches of the bottom of the enclosure would meet the minimum requirements.

#### One-permanent-opening method:

The minimum free area of 1 square inches per 3,000 Btu/h of total input rating of all appliances in the enclosure.

( Total BTU Input of all Appliances / 3,000 = Total Free Area Required ).

(120,000 BTU / 3,000) x 1 in2 = 40 in2 minimum free utilizing one or more openings

Utilizing metal louvers requires (40 in2/.75 = 53.3 in2) to obtain 53.3 in2 of free area. Confirmation: 75% of 53.3 in2 = .75 x 53.3 in2 = 40 in2

One opening 53.3 in2 within 12 inches of the top of the enclosure would meet the minimum requirements.

### Combination Indoor & Outdoor Combustion Air:

For example we have 3,000 ft3 of indoor combustion air available from the same story

As calculated above we will need one opening 160 in2 within 12 inches of the top and one opening 160 in2 within 12 inches of the bottom of the enclosure to meet part of the combustion air requirements.

Without the indoor air, utilizing only the one-permanent-opening method as shown above we would need one opening 53.3 in2, however, we can reduce this because we are utilizing some indoor air for combustion.

The reduction factor equals the calculated total free area based on the outdoor combustion air section x (1 - available volume/required volume).

53.3 in2 x (1 - (3,000ft3/6,000ft3)) = 26.65in2

Therefore, we need the two 160in2 openings communicating with the indoor air and one opening 26.65 in2 within 12 inches of the top of the enclosure in order to meet the minimum requirements.