Document jyjR6npwvg8OaDwJMyGDLXBg5

American Society of Heating and Ventilating Engineers Guide When equilibrium is established, the amount of heat flowing through any comp-t part of a construction is the same for each square foot of area. as* Therefore, where U [ 80 - (-10) 1 = 1.65 (80 - 65) U is the transmittance of the insulated roof. Solving the equation, V = 0.275. The resistance of the insulated roof = 1 0.275 =3.64. The resistance of the uninsulated roof = = 1.39. The resistance of the insulation = 3.64 -- 1.39 = 2.25. Resistance per inch of insulation = 'Q1ljjj' =3.0. Since a resistance of 2.25 is required, and 1 in. of insulation has a resistance of 3, one iori will be sufficient to prevent condensation. The same result might have been obtained by selecting an insulated 4-in. concrete sla' having a U of less than 0.275 from Table 11, Chapter 5. This 4-in. concrete slab 1-in. rigid insulation has a if of 0.23 which is safe. 2 What inside dry-bulb temperatures are usually assumed for: (a) hom; (b) schools, (c) public buildings? Referring to Table 1: a. 70 to 72 F. j; b. Temperature varies from 55 to 75 F, depending on the room. Classrooms, for instance^ are usually specified as 70 to 72 F. c. 68 to 72 F. . 3 How is the outside temperature selected for use in computing heat losses! The outside temperature used in computing heat losses is generally taken from 10 to 15F higher than the lowest recorded temperature as reported by the Weather Bureau dunsf the preceding 10 years for the locality in which the heating system is to be installed,'. In some cases where the lowest recorded temperature is extremely unusual, the design] temperature is taken even higher, than 15 F above the lowest recorded temperature. 4 What are the effects of wind movement on the heating load? . Wind movement increases the heat transmission of walls, glass, and roof; it affects poor walls to a much greater extent than good walls. . Wind movement materially increases the infiltration (inleakage) of cold air thfbugli the cracks around doors and windows, and even through the building materials them selves if such materials are at all porous. 5 Calculate the heat given off by eighteen 200-watt lamps. 200 X 18 X 3.415 = 12,294 Btu per hour, 6 A two-story, six room, frame house, 28-ft by 30-ft foundation, has the following proportions: Area of outside walls, 1992 sq ft. Area of glass, 333 sq ft. Area of outside floors, 54 sq ft. Cracks around windows, 440 ft. Cracks around doors, 54 ft. Area of second floor ceiling, 783 sq ft. Volume, first and second floors, 13,010 cu ft. Ceilings, 9 ft high. 142 i Chapter 7--Heating Load -erature for the heating season is --34 F, and the required The |0ininJUn *the 30-in. level is 70 F. The average number of degree inside tenit*^8* season is 7851, and the average wind velocity is 10 mpb, Zys for a heating sea TOpa0h5peretwbra,wJaeUn8S`d wocoodnssitdrumctgeodnotfh2e-ionu.tbsiyd4e,-iann. dstwuodosdwlaiththwaonoddpslahsetaetrhoinngt,hbeuiinldsiindge! Windows are single glass, wood, without weatherstrips. The sI.eocofnwdofolodosrhcinegilliensgoisn mwoeotadlslatrtiphsawndithplraasftteerr,swexitphoosuedt .anTahtetiacreflaooorf. thTehreorooftosf 2(a0)pVerfcoernwt garlelsa;te(br )thUanfotrheglaarsesa; o(cf)thUefcoerilisnegc.onSdeflleocotrvcaeluielisngfo; r(dth)eVfoflolorwroinogf- (e) V for ceiling and roof combmed; ff) air leakage, cubic feet per hour per fW of window crack; (g) air leakage, cubic feet per hour per foot of door crack. o._0.25 (Table 5, Chapter 5). d. 1.13 (TaBle~13,_Chapter-5).---- --------- 0.69 (Table 8, Chapter 5). 0.48 (Table 12, Chapter 5). 0.236 (Equation 6, Chapter 5). 21.4 (Table 2, Chapter 6). 42.8, which is double the window leakage. 7the UvasriniogutshceodnasttaruocftiQonuse,sationnd 6s,hcoawlctuhleatpeetrhceenmtaagxeimouf mtheBttuotlaolsshpSetr hwohuircfho*r last through each construction described. 1 ",cn ,s lost uw--o- Assume 2 per cent rise in temperature for each foot in height. The average temperature mil be 72.8 F for walls, doors, and windows, and 79.1 F for the second'floor ceiling a. Outside walls b. Glass c. Doors d. Second floor ceiling e. Air leakage, windows /. Air leakage, doors 46,200 Btu loss 34,950 Btu loss 5,670 Btu loss 17,840 Btu loss 15,750 Btu loss 3,865 Btu loss '37.2 per cent of total 28.1 per cent of total 4.6 per cent of total 14.3 per cent of total 12.7 per cent of total 3.1 per cent of total Totai 124,275 Btu loss 100.0 per cent of total 8 For the house in Question 6, place I-in. insulation in the outside walls and second floor ceiling; k for insulation = 0.34. Use weatherstrip on doors and windows, and double glass on the windows; Ca -- 0.55. Calculate or select the following values: (a) V for walls; (b) U for glass; (c) U for second floor ceiling; (d) U for combination of ceiling and roof; (e) Air leakage, cubic feet per hour per foot of door crack; (f) air leakage, cubic feet per hour per foot of window crack. 0.144. 0.55. 0.23. 0.13. 15.5. 31.0. 9 Calculate the maximum Btu loss per hour and show the percentage loss by each channel for the house as insulated in Question 8. ?* Outside walls Glass - Doors 4* Ceiling f^ windows / Air leakage, doors 26,650 Btu loss 17,000 Btu loss 5,670 Btu loss 10,070 Btu loss 11,400 Btu loss 2,795 Btu loss 36.2 per cent of total 23.1 per cent of total 7.7 per cent of total 13.7 per cent of total 15.5 per cent of total 3.8 per cent of total Total 73,585 Btu loss 100.0 per cent of total 143