Document 6BDbam5757rBbN9yd4jJLpEa1

American Society of Heating and Ventilating Engineers Guide, 1926-27 above breathing line. Allowing 2 per cent per foot above 5 ft., or 2 X 15 = 30 per cent, makes the under roof temperature = 1.30 X 60 = 78 deg. fahr. Floor: The 5 in. concrete floor is laid on the ground, and hence there is only one surface coefficient K, = 1'.3 (Table 3). U = ----i------- = 0.717 --1.I3-0---^h --8 The air temperature at floor level = 60 -- 5 = 55 deg. fahr. Windows and Doors: Wood sash and doors with single thickness of glass. Take coefficient U for glass as 1.13 B.t.u. per sq. ft. per degree pier hour for heat transmission (Table 6). Doors are solid wood 1% in. thick and coefficient U = 0.37 B.t.u. per sq. ft. per degree per hour (Table 6). Infiltration: Window crack assumed xg in. and doors at in. By Table 14 (Part I) for a 10 mile wind velocity the leakage pier foot of crack is 85 c.f. hr. for a plain window. The heat equivalent per hour, per degree is 85 X .075 X 0.24 = 1.53 B.t.u. and allowing for an 11 mile wind the factor becomes 1.53 X = 1.68 (see preceding note). Allow twice this for door crack or 2 X 3 -68 = 3.36. Building Material. Exposure Concrete and Tile__ N N Doors ................. N Crack yt'................... N - Calculation Sheet Entire Building (See Fig. 7) Coeffic. Area Trans. Width Height Sq. Ft. and Temp. in Ft. in Ft. or Lin. Infill. Diff. Ft. M X 50 8^ 213 0.29 74.0 50 16 656 0.29 59.4 12 12 144 0.37 56 1 pair doors 60 3.56 56 Net B.t.u. 4,570 11,300 2,980 11,900 Exposure Factor 1.15 2.15 1.15 H* X 1.15 Total B.t.u. 5,250 13,000 3.430 6.850 ' 28.530 Concrete and Tile__ Glass............................ Crack >6'........... ....... W w w 120 16 15 X 4 9 Double Hung 1380 0.29 540 1.13 450 1.78 59.4 59.4 59.4 22.900 36.200 47.500 1.15 1.15 X 1.15 26,300 41.600 27,300 Windows (15) 95,200 South Wall................. Same as N See above 30,750 M* 24,800 East Wall................... Same as-W Roof 3* Concrete See above 106,600 H* . 82,850 and Slag............ No Ceiling 52.5 120 6300 0.60 74 280,000 None 280,000 Floor 5' Stone Con- . crete.... .................... On Dirt 50 120 6000 0.717 5 21.510 None 21,510 Grand total of heat required for building in B.t.u. per hour at 4- 4 9 with 11-mile Southwest wind.. 532,890 Notes--(1) This building has no partitions and whatever air enters through the cracks on the wind ward side must leave through the cracks on the leeward side. Therefore, only one-half of the total crack will be used in computing infiltration for each side and each end of building. (2) An exposure allowance of 15 per cent is also to be added to the wall and glass transmission losses and to the infiltration losses on the two adjacent sides of the building most nearly facing the prevailing wind as stated in paragraph 36. (3) It is also possible to compute the heat required to take care of infiltration on the basis of M of an air change per hour-as given in Table 13 for a factory with minimum conditions. Volume =* 50 X 120 X 20 (mean height) = 120.000 cu. ft. and heat required per hour is 120,000 X M X 0.075 X 0.24 X 59.4 = 64,200 B.t.u. Based on infiltration through one-half the total crackage in all walls, the heat to be supplied per hour is from preceding table, 6.850 + 27,300 + 5,950 + 23,750 - 63,850 B.t.u. This value based on crackage should be used, but if building is to be heated intermittently, not less than one air change per hour should be allowed. 32 Chapter II HEATING BY RADIATION CALCULATING RADIATION RADIATION can be classified as direct, semi-direct, and indirect, and is usually made of pipe or cast iron; when it is made of pipe it is termed pipe coil, and when made of cast iron it is termed column, wall, semi-indirect, or indirect radiation. The unit of measure'in figuring radiation is the square foot of heating surface, which is the external surface. The amount of heat a square foot of heating surface (radiation) will give off depends upon the temperature of the heating medium (steam; or hot water), the temperature of the surrounding air, and the velocity at which the air passes over same. Tables 18 to 24 on succeeding pages indicate the number of B.t.u. a given size column or wall radiator will transmit in 1 hr. with steam as the heating medium and Tables 26 to 30 give similar data with hot water as the heating medium. The ordinary practice in calculating the amount of radiation of various kinds to meet a variety of conditions will be briefly stated. To determine the amount of direct radiation to heat a room, figure all of the heat losses, add the proper amount for exposure, and refer to Tables 18 to 30 to find the proper size radiator. If a radiator of more than 20 sections is required, multiply the value, B.t.u. pier intermediate section, for the particular radiator, Tables 18-24, by the number of additional sections above 20 and add this amount to the value, Total B.t.u. per hour, for the 20 section radiator. This will give the total B.t.u. per hour for the required radiator. Example.--What is the total B.t.u. per hour for a 30-section .32 in., single column radiator? Solution.--626 B.t.u. X 10 Sections = 6260 B.t.u. 6260 B.t.u. -f- 12,875 B.t.u. = 19,135 B.t.u. Total per hour for 30-section radiator. The values, B.t.u. per square feet of intermediate section and B.t.u. per square feet of end surface, are given on the tables to show the rela tionship between the two. The greater exposure of the end surface on the radiator, obviously will give a greater emission per square feet of surface. To determine the amount of semi-indirect~(sometimes termed directindirect), radiation to heat a room, figure all the heat losses, adding the Prepared especially for The Guide by R. V. Frost, Norristown, Pa. 33