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American Society of Heating and Ventilating Engineers Guide, 1937 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 12 at the end of this chapter. HEAT LOSS COMPUTATION EXAMPLE 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. Hence the base temperature = (- 6 + 10) = + 4 F. 4. Direction of prevailing wind (during Dec., Jan., Feb.)_____ -Northwest 5. Breathing-line temperature (5 ft from floor)________________ ______60 F 6. Inside air temperature at roof: >. The air temperature just 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 11 =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 = 63.6F. 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. 152 Chapter 7--Heating Load Floor--5-in. stone concrete on 3-in. cinder concrete on dirt. Doors--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)U = 0.34 Roof--(Table 11, Chapter 5,-Roofs 2A and 3A)U = 0.77 Floor--(Table 10, Chapter 5, Floors 5A and 6A)J______ U = 0.63 Doors--(Table 13B, Chapter 5)--------------------- _______________ U = 0.46 Windows--(Table 13A, Chapter 5)____ V = 1.13 12. Infiltration Coefficients: Windows--Average windows, non-weatherstripped, JC6-in. crack and %4-in. 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 cracks Doors--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 Past of Building North Wall: Brick. H-in. plaster Doors (2-in. wood)_________ H in. Crack. -- West Wall: Brick. K4n. plaster. Glass (Single)_______________ % in. Crack. . . South Wall East Wall Roof. 3-in. concrete and slagsurfaced built-up roofing Floor, 5-in. stone concrete on 34n. tinder concrete_________ Width in Feet Height in Feet Net Sur face Area or Crack Length 60 16 12 12 1 pair doom 656 144 60 . 120 16 15x4 9 Double Hung Windowre (15) 1380 540 450 Same as North Wall Same as West Wall 50 120 6000 50 120 6000 Coeffi cient 0.34 0.46 0.90 ' 0.34 1.13 0.45 0.77 0.63 Temp. Diff. Total Btu 59.6 57.2 57.2 59.6 60.2 60.2 69.2 5b 13.293 3.789 1.544* 27.964, 36,734 6,095* 18.626 70.793 319,704 18.900 Grand Total of heat reauired for hnildin? In Rtn nw hour 517.442 building has no partitions and whatever air enters through the cracks on the windward side must tracks on the leeward side. Therefore, only one-half of the total crack will be used in mpuung infiltration for each side and each end of building. temperature differential is commonly assumed to exist between the air on one side of a large' "or laid on the ground and the ground. 153