Document 315jEQR3yebbRoEYdapy3mZ3

American Society of Heating and Ventilating Engineers Guide, 1936 surface resistance. This can be accomplished by increasing the velocity of air passing over the surface, or by increasing the over-all resistance of the wall or roof by installing a sufficient thickness of insulation. . The latter method is generally used, and the thickness of insulation is determined by ascertaining the amount of resistance to be added to increase the temperature of the interior surface above the dew-point temperature for the maximum conditions involved. This in turn is based on the fundamental principle that the drop in temperature is proportional to the resistance. See Question Hat the end of this chapter, EXAMPLES OF HEAT LOSS COMPUTATIONS Fig. 1. Elevation of Factory Building 1. Location--------------- :.................... .................................... ................................. Philadelphia, Pa. 2. Lowest outside temperature. (Table 2)_____________ ___________________ -- 6F 3. Base temperature: In this example a design temperature 10 F above lowest on record instead of 15 F is used.xyHence the base temperature = (- 6 + 10) = + 4 F. 4. Direction of prevailing wind (during Dec., Jan., Feb.)_____________Northwest i 5. Breathing-line temperature (5 ft from floor)_________________________;;_.60 F 6. Inside air temperature at roof: The air temperaturejust below roof is higher than at the breathing line. Height of roof is 16 ft, or it is 16 -- 5 = 11 ft above breathing line. Allowing 2 per cent per foot above 5 ft, or 2 X H = 22 per cent, makes the tem perature of the air under the roof = 1.22 X 60 = 73.2 F. .\ 7. Inside temperature at walls: The air temperature at the mean height of the walls is greater than at the breathing line. The mean height of the walls is 8 ft and allowing 2 per cent per foot above 5 ft, the average mean temperature of the walls is 1.06 X 60 = 63.6 F. By similar assumptions and calculations, the mean temperature of the glass will be found to be 64.2 F and that of the doors 61.2 F. T 8. Average wind velocity (Table 2)__ ...:................ .................................. ;.11.0 mph 9. Over-all dimensions (See Fig. 1).,............................. ............... ................ 120 x 50 x 16 ft 10. Construction: Walls--12-in. brick, with J^-in. plaster applied directly to inside surface: Roof--3-in. stone concrete and built-up roofing. 152 Chapter 7--Heating Load pioor--5-in. stone concrete on 3-in. cinder concrete on dirt. Poors--One 12 ft x 12 ft wood door (2 in. thick) at each end. Windows__Fifteen, 9 ft x 4 ft single glass double-hung windows on each side. 11. Transmission coefficients: Walls--(Table 3, Chapter 5, Wall 2B)._,,............................. gooj_(Table 11, Chapter. 5, Roofs 2A and 3A)..,,---------- U = 0.34 TJ = 0.77 floor--(Table 10, Chapter 5, Floors 5A and 6A)........... __ U = 0.63 Doors--(labie iod, \_ndptci v- -....................................... --............. TJ = 0.46 Windows--(Table 13A, Chapter 5)------------------ ------------ _......... U = 1.13 12. Infiltration Coefficients:, Windows--Average windows, non-weatherstripped, crack and He-ia clearance. The leakage per foot of crack for an 11-mile wind velocity is 25.0 cfh. (Determined by interpolation of Table 2, Chapter 6.) The heat equivalent per hour per degree per foot of crack is taken from Chapter 6. 25.0 X 0.018 = 0.45 Btu per deg Fahr per foot of crack. Poors__Assume infiltration loss through door crack twice that of windows or 2 X 0.45 = 0.90 Btu per deg Fahr per foot of crack. Walls--As shown by Table 1, Chapter 6, a plastered wall allows so little infiltration that in this problem it may be neglected. 13. Calculations: See calculation sheet, Table 3. Table 3. Calculation Sheet Showing Method of Estimating Heat Losses of Building Shown, in Fig. 1 Part of Building Width in Feet Height in Feet Net Sur face Area or Crack Length Coeffi cient North Wall: Brick, H-in. plaster............. ... Doors (2-in. wood)_________ H in. Crack.. ........................ West Wall: Brick, H-in. plaster................ Glass (Single)...... ......... -......... % in. Crack..... ......................... South Wall..................................... East Wall. ........................... Roof, 3-in. concrete and slag surfaced built-up roofing.^.__ Floor, 5-in. stone concrete on 3-in. cinder concrete 50 16 12 12 1 pair doors 656 , 144 60 120 16 15x4 | 9 Double Hung Windows (15) 1380 540 450 Same as North Wall Same as West Wall 50 120 6000 50 120 6000 0.34 0.46 0.90 0.34 . 1.13 0.45 0.77 0.6 Temp. Diff. 59.6 57.2 57.2 59.6 60.2 60.2 a 69.2 5b Total Btu 13,293 3,789 1,544a 27,964 36,734 . 6,095a 18,626 70,793 319,704 18.900 Grand Total of heat required for building In Btu per hour. 517,442 This building has no partitions and whatever air enters through the cracks on the windward 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. bA 5 F temperature differential is commonly assumed to exist between the air on one side of a largefloor laid on the ground and the ground. 153