Document Y9vNQzn8MMV9wyaYowXzg5bE

American Society of Heating and Ventilating Engineers Guide, 1930 Leader diameter = 11.4, say 12 in. Stack area = 0.7 X 90 = 63 sq. in. = say 5 X 12 in. Register area = 90 sq. in. net area. Gross area = 1.6 X net area = 12 X 12 or 12 X 14 in. In like manner the leaders, stacks and registers are calculated for each room in the house. Leaders, Stacks and Registers. (Code Method. See Art. 3, Sec. 1, 2, 3) Living Room (Glass = 90, Net wall = 405, Cubic contents = 2405) T, /90 , 405 , 2405 ... Leader = {^ + 60 + W ) 9 = 155 ** " Register, same as Direct Method. Owner's Room (Glass = 68, Net wall = 394, Cubic contents = 2275) ., (68 . 394 , 2275 \ ,, ' . Leader = (^ + 50 + W ) 6 = 90 " m` Stack and Register, same as Direct Method. Assuming all air recirculated, the minimum furnace for the plant will be: Grate Area = 0.0034 X 132,370 s= 450 sq. in. = 24 in. diam. at 175 deg. register temperature. (7) Grate Area = 0.0040 X 132,370 = 530 sq. in. = 26 in. diam. at 160 deg. register temperature. (8) (10) If provision shall be made for certain outside air circulation, then increase the building heat loss by, say 25 per cent and obtain by equation (7) a 27-in. grate and by equations .(8) and (10) a 29-in. grate. Table 1. Summary of Data Applied to Warm Air Research Residence Rooms From Chapter 2 on Heat Losses from Buildings B.t.u. Heat Losses H Leader Area Sq. In. Stack Area Sq. In. 0.7 X LA Leader Diameter Inches Stack Size Net Register Size Gross First Floor Dining______ Hal! and stair Second Floor Owners........... S. W. Bed___ Bath................ N. Bed........... Third Floor E. Bed. ........ W. Bed.......... 17250 6810 2300 9210 25710 12570 15030 9800 2450 14800 8220 8220 = 0.0091/ 155 61 21 83 230 113 = 0.00617 90 59 15 89 = 0.005/7 41 41 63 41 10 62 29 29 14 9 8 11 or 12 Two 12 12 11 or 12 5 X 12 9 3y2 X 12 8 3 X 10 11 or 12 5 X 12 8 3 X 10 8 3 X 10 14 X 18 8 X 12 8 X 10 12 X 14 12 X 14 12 X 14 8X12 8 X 10 12 X 14 8 X 10 8 X 10 140 Chapter 5--Heating with Warm Air Furnaces by Gravity STANDARD CODE REGULATING THE INSTALLATION OF GRAVITY WARM AIR FURNACES IN RESIDENCES3 SIXTH EDITION March 1, 1929 ARTICLE No. 1.--Meaning of the Term "Gravity Warm Air Furnace Heating System" Gravity Warm Air Heating Systems, to which this code refers, shall consist of one or more warm-air furnaces, enclosed within casings, together with necessary appur tenances thereto, consisting of warm-air pipes and fittings; cold air or recirculating pipes; ducts, boxes and fittings; smoke-pipes and fittings; registers, borders, faces and grilles; the same being intended for heating buildings in which they may be installed. ARTICLE No. 2.--Certified Measurements Certified measurements on warm-air furnaces together with the name and number of that furnace, will be issued by authority of the National Warm Air Heating A ssociation, when, if and as, the grate areas and heating surfaces have been accurately measured and approved by the Research Advisory Committee. ARTICLE No. 3.--Method for Determining Size of Warm Air Pipes, Wall Stacks and Furnaces for Use in Residences Method for Determining Sizes of Basement Warm Air Pipes Section 1. Each First-Floor Room: Divide square feet of glass by 12 Divide square feet of net outside wall by factor in Table A Divide cubic contents by 800 Add together the above and multiply by 9 The result is the area of the basement pipe in sq. in. Stated as an equation, this is: /The sum of: ) Glass (sq. ft.) (Note 1) -s- 12 | X.9 = area of basement,pipe ) Net Wall (sq. ft.) (Note 2) -f* factor in Table A (Cubic Contents -f- 800 Section 2. Each SecondvFJoor Room: Divide square feet of glass by 12 Divide square feet of net outside wall by factor in Table A Divide cubic contents by 800 Add together the above and multiply by 6 The result is the area of the basement pipe in sq. in. (See Section 9. c) Stated as an equation, this is: The sum of: Glass (sq. ft.) (Note 7) + 12 | X 6 = area of basement duct Net Wall (sq. ft.) (Note 2) ~ factor in Table A Cubic Contents -r- 800 Section 3. Each Third-Floor Room: Divide square feet of glass by 12 Divide square feet of net outside wall by factor in Table A Divide cubic contents by 800 Add together the above and multiply by 5 The result is the area of the basement pipe in sq. in. Stated as an equation, this is: The sum of: Glass (sq. ft.) (Note 1) + 12 | X 5 = area of basement duct Net Wall (sq. ft.) (Note 2) -s- factor in Table A Cubic Contents -4- 800 *This Code .is approved and issued by authority of the National Warm Air Heating Association, The American Society of Heating and Ventilating Engineers and the National Association Sheet Metal Contractors. First edition. October 1, 1922; 2nd edition. February 1, 1923; 3rd edition. June 1, 1924; 4th edition. May 1, 1927; 5th edition, March 1, 1928; 6th edition. March 1. 1929. 141