Document MGdLrd8zpjqdVgbrE141pgLZj

Heating Ventilating Air Conditioning Guide 1938 tures of the various parts of a wall are controlled by the type and amount of insulation used and the vapor densities in the corresponding sections are controlled by the type of vapor barriers installed. The transmission of hejat and vapor through a wall should be considered together, and- in most cases the . proper .combination of insulation and vapor barriers will eliminate the possibilities of condensation within walls. A consideration often overlooked in problems of condensation within walls is that a vapor barrier should be placed on the warm side and not on the cold side of a wall. HEAT LOSS COMPUTATION EXAMPLE 1. Location.;-------------------------------------------------------------------- ........------- Philadelphia, Pa. 2. Lowest outside temperature, (Table 2)------- ---!--L;-- ------- _---------- --. --. 6 F 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 = 1 (- 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 temperdture \ast 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 ll = 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. 8. Average wind velocity (Table 2)-------------------------- .---------1--------- i------ 11.0 mph 9. Over-all dimensions (See Fig. 1):_;--....................... ...................... --120 x 50 x 16 ft 10. Construction: Walls--12-in. brick, with H-in. plaster applied directly to inside surface. Roof--3-in. stone concrete and built-up roofing. 142 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)U = 0.63 Doors--(Table 13B, Chapter 5)^_________________ ....... ....... U -- 0.46i Windows--(Table 13A, Chapter 5).__U = 1.13 , 12. Infiltration Coefficients: ' * Windows--Average windows, non-weatherstripped, Jf6-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 Fahrenheit per foot of crack. Doors--Assume infiltration loss through door crack twice that of windows., or 2 X 0.45 = 0.90 Btu per deg.Fahrenheit 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 North Wall: . Brick. H-in. plaster_________ Doors (2-in. wood)..... K in. Crack. _ __________ West Wall: Brick. H~in. plaster Glass fSineie) .. .. J6 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.________ Width IN Feet Height in Feet Net Sur face Area or Crack Length Coeffi cient 50 16 12 12 .1 pair doors 656 \ 144 60 0.34 0.46 0.90 120 16 15x4 9 Double Hung Windov79 (15) 1380 540 450 Same as North Wall Same as West Wall . 0.34 : 1.13 0.45 50 120 6000 0.77 50 120 6000 0.63 Grand Total of heat reauired for hiiiMinv In Rt.n nw hour Temp. Diff. Total Btu 59.6 57.2 57.2' 13.293 3,789 1,544* 59.6 60*2 60.2 27,964 36,734 6,095* 18,626 70.793 69.2 319.704 5b 18,900 517,442 , ha3 ? partitions and whatever air enters through the cracks on the windward aide must ism urough 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. temperature differential is commonly assumed to exist between the air on one side of a large noor lard on the ground and the ground. 143