Document 2Jqyg6V2p2Ga1xzqEKEEELgOp

324 CHAPTER 13 1956 Guide Solution: From Table 3, the recommended ventilation rateis 15 cfm per person. . Total necessary = 85 X 15 = 1275 cfm or 76,500 cu ft per hr. As the room volume is 40,000 cu ft; the air changes per hour will be 76,500/40,000 =* 1.91 which is more.than one air change. js .. Estimated Time of Maximum Cooling Load: ' J For this job, judgment indicates that the'roof will make the greatest single con tribution to the cooling load. Hence, the time Of Maximum.cooling load probably will be the time of maximum beat gain through the roof. , From Table 9 the maxi mum temperature differential for a 2 in. gypsum roof of medium weight construction is 54 deg at 4:00,p.m., and 53 deg at 3:00 p.m." Examination of Table 13 (40 deg N Latitude) shows that solar heat gain through glass on the south wall is 18.Btu per (hr) (sq ft) at 4:00 p.m., and 42 Btu at 3:00 p.m. This indicates that the maximum cooling load occurs at approximately 3:00 p.m. Therefore make load calculations at 3:00 p.m. sun time. (This may be slightly different from 3:00 p.m. local time.) In some cases, there would be no clear-cut evidence of this nature,1 and consequently, it would be necessary to estimate the load for several successive times, and then to select the. maximum. .. . ., ... Heat Gain Through Outer Wall and Roof Areas: ... . ...... .: From Table 10 the temperature differential for the south wall (8 in: concrete block with 4 in. brick veneer) may be about the same as a 12 in. brick which is 6 deg at 3:00 p.m. for a dark colored wall. From the same table, the temperature differential for the east wall (8 in. concrete block with plaster) will be 11 deg at 3:00 p.m. for a light colored wall (interpolating between 2:00 and 4:00 p.m.). Likewise, the tem perature differential for the north exposed wall (8 in. concrete block plus plaster) will be 3 deg at 3:00 p.m. (by interpolation) for a light wall. The party wall of 13 in. brick on the west side and part of the north side may be treated as if it were an outside wall in the shade which has a temperature differential (from Table 10) of;2 deg. v For the door in.north wall estimate V = 0.59 from Chapter 9, Table 10, No. 5A. The outdoor temperature at 3:00 p.m. is 95 F. Neglect time lag and any decrement factor. The temperature differential is (<P -- <I) = 95 -- 80 = IS deg. The tabula tion of the preceding values at 3:00 p.m. is given in the following table: Section Net Area Sq Ft Temperature Differential F Deg Heat 7 Transmission Coefficient {V) Heat Flow . Rate per Hour Btu Roof South Wall East Wall North Exposed Wall West A North Party Wall Door in North Wall 4000 405* 765" 170* 1065* 35 53 6 11 3 2 15 * Calculated from gross wall area, less windows and doom. 0.34 0.41 0.62 0.62 0.26 0.69 72,000 995 4,380 265 550 310 78,600 ----------------------- Heat Gain Through Glass Areas In computing the load for 3:00 p.m., only the south windows and doors will be exposed to direct sunlight: Tables 13 and 14 will give the total heat gain from the glass areas. The window reveals will shade the south windows; the fraction of the window area receiving direct radiation is obtained from Equation 5 by substituting values as follows: r, = s/1 = 4/60; r, = 4/36; 0 = 45.5 deg, tan 0 = 1.02 y = 74 deg, tan y = 3.487, cos y = 0.276 0* = 1- <K<H) - 36 <3 487> + (a) (si) (1-02) (3.487) 0.276 0.462. The south doors will be considered entirely sunlit. The outdoor air temperature Cooling' Load 325 is;95' F at 3:00 p:mv. From Table 24-the inside Venetian1 blind factor is 0.65. The instantaneous heat gains due to transmitted direct and diffuse solar radiation, and from convection and radiation gain, are found in Tables 13 and 14 as listed below for the south facing-doors and windows, the-north: facing windows and the J glass doors in the east wall., The gain through the solid portion of the east doors may be approximated by use of Fig. 3, since the wood panels have little heat capacity. From Table 4. the, diffuse radiation value is taken as 18 Btu.per (hr)(sq ft) from which i,, ajjfm is found to be 98.2 for a = 0.7 and -- '4.0. From Fig. 3, q = 27.0 Btu per (hr)(sq ft). These heat gains are itemized in the following table (note that there:are no corrections to Table 14 values since table is based on 80 F tetriperfiture, but Fig. 3is bUsed on 75F room temperature). Location South Windows South Doors '1 East Doors/Glass IWood North Windows Total * Area ` Frac tion Sunlit Sq Ft Shade Factor Trans Solar Gain . Btu/ (hr) (sq ft) Cork Conv and Rad Gain from 75F to 80F Btu/ Indoor (hr) Temper (8Q FT) ature Btu/(hr) (sq ft) Total Gain Btu/ (hr) (sq ft) Total Gain Btu/hb 60 . 0.462 0.65 . 13 19 ` 35 ' 1.00 42 19 18 ' 14 17 18 27 - 30 15 17 32 1920 61 2136 31 '560 ' 22 395 32 960 5970 In some jobs it would be desirable to increase (or decrease) the instantaneous radi ation heat gain by a load-lag factor. The reason for not doing so in this case is that the solar gain is of a low magnitude, and reference to the table indicates that 0.8 of the previous hour would not affect the results materially. Heal Gain from Ventilation and Infiltration: Since the desired outdoor air rate 1275 cfm is greater than one air change per hgaoiunr., it will be satisfactory for determining the ventilation component of the heat Window infiltration can be taken as negligible since the windows do not open. Door infiltration requires some judgment. Assume that for each person passing through the double doors, the infiltration will be 100 cu ft of outdoor air, see Chapter 10, Table 3. Assume that the outside doors will be used at the rate of 10 persons per .hour and the inside doors at the rate of 30 persons per hour. Total infiltration will then be 40 X 100 = 4000 cfh or 67 cfm. The design rate of entry of outside air is then: The sensible, latent and total loads are determined from Equations 7, 8, and 9, respectively, at 3:0Q p.m. (Table 8) U = 95, L -- 80, Wo = 0.0169, Wi = 0.0098. All the air entering the room as infiltration becomes a part of the space load. Infiltration (see Equations 7, 8, and 9): q. = ; 67 X 1.08 (95-80) = 1085.Btuh, sensible. 9. = _67 X 4840 (0.0169-6.0098) = 2300 Btuh, latent. ' <?t = q, + 9c = 1085 + 2300 = 3385 Btuh, total. Ventilation Air Taken through Cooling Unit Which Becomes a Part of the Space load (see Equations 10 and 11): . q.i = 1275 X 1-08 (95-80) (0.15) = 3,100 Btuh, sensible. ' ffei = 1275 X 4840 (0.0169-0.0098) (015) = 6,570 Btuh, latent.