Document LoLyojJ59pbZexvaD6bNwQYq

318 CHAPTER 15 1949 Guide > South wall construction:^ in. concrete block, 4 in. brick veneer,.$ in. plaster,, on walls. (Table 8, Chapter 6, No. 92B, U = 0.41.) .: East wall and outside north wall construction: 8 in. concrete block, painted white, J in. plaster on walls. (Table 7, Chapter 6, No. 82B, U = 0.52.) . West wall and adjoining north party wall construction: 13 in. solid brick, no plaster: V-m + 7 + m ^ = 0-263. Use (7 = 0.26. Hoof construction: 2 in. flat roof deck of 2 in. gypsum fiber concrete on gypsumboard surfaced with built-up roofing. -- (Table 16, U = 0.34 for summer.) Floor construction: 4 in. concrete on ground. Window: 3 ft x 5 ft, non-opening type, with medium colored Venetian blinds for windows on south wall. Approximately 4 in. reveal on all'windows. Front doors: Two 2 ft-6 in. x 7 ft (glass panels). Side doors: Two 2 ft-6 in. x 7 ft (J glass panels). Rear doors: Two 2 ft-6 in. x 7 ft (glass panels). . Outside design conditions: Maximum dry bulb 95 F, wet bulb 78 F; W0 = 0.0169 lbs vapor per lb dry air; ho. -- 41.38 Btu per lb dry air. Indoor design conditions: Dry bulb 80 F, wet bulb 65 F; W\ = 0.0098 lb vapor per lb dry air; h\ = 29.95 Btu per lb dry air. Occupancy: 85 office workers. ' Lights: 12,000 watts, fluorescent; 4000 watts tungsten. Fan motor: 7 hp. Conditioning equipment to be located in adjoining structure to north. - Find: total, sensible, and latent maximum cooling loads and required air quantity through conditioning equipment. Solution: From Table 4, the recommended ventilation rate, is 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. Estimated Time of Maximum Cooling Load: 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 heat gain through the roof. From Table 14 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 18 (40 deg N Latitude) shows that solar heat gain through glass on the south wall is 19 Btu per (hr) (sq ft) at 4:00 PM. and 42 Btu at 3:00 PM. This indicates that the maximum cooling load occurs at approximately 3:00 PM. Therefore make load calculations at 3:00 P.M. suntime. (This may be slightly different than 3:00 PM. local time.) In some cases, there would be no clear-cut evidence of this nature and consequently it would be necessary to estimate the load for several successive times and then to select the maximum. Heal Gain Through Outer Wall and Roof Areas: From Table 15 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 whichis 6deg at 3:00 PM. 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 PM. for a light colored wall (interpolating between 2:00 and 4:00 PM:). Likewise, the tem perature differential forthe north exposed wall (8 in. concrete block plus plaster) will be 3 deg at 3:00 PM. (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 15) of 2 deg. ; . For the door in North Wall estimate U -- 0.59 from Chapter 6, Table 9, No. 5A. The outdoor temperature at 3:00 P.M. is 95 F. Neglect time lag and any decrement factor. The temperature differential is (fp -- ti) = 95 -- 80 = i5 deg. . The tabula tion of the preceding values at 3 p.M.- is given jn tfie.following table; Cooling^ Load 319 --- ' ------ Section Roof South Wall -'.-.East Wall ( North Exposed Wall . West & North Party Wall Door in North Wall Net Area Bq.Ft Trupebature Differential F Deg Transmission Coefficient (U) ' ' Heat Flow Rate feb Houb ' Btu 4000 405* 765* 170* 1065* 35 53 . 0.34 6 0.41 11 0,52 . 3 . 0.52. 2 0.26 15 0.59 72,000 995 4,380 265 550 310 . 78,500 Calculated from gross wall area less windows and doors.' 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. Table 18 and Equation 10 will give the total heat gain from the glass areas. The window reveals will shade the south windows slightly; the fraction of window area receiving direct radiation is obtained from Equation 11 by substituting values as follows: . n -- s/l -- 4/60; r* = 4/36; 0 = 45.5 deg, tan 0 ~ 1.02. y = 16 deg, tan y = 0.287. (i)UJ= 1-^ (1.02) - 346 (0.287) + ! (i:02) (0.287) = 0,902. The south doors will be considered entirely sunlit. The outdoor air temperature is 95 F at 3:00 P.M.. From Table 23 the inside Venetian blind factor is taken as 0.74. Referring to Table 18, the instantaneous heat gain due to solar and sky radia tion for 40 deg N. latitude for south exposure at 3:00 P.M. is read as 42 Btu per sq ft. These figures are tabulated below. The radiation gain for south windows is 60 X 0.902 X 0.74 X 42 = 1680 .Btu per hr. The normal heat transmission 60 X 1.04 (95-- 80) = 940. The.sum of these heat gains 2620. Btu per hr is.totaled in last column. The remaining doors and windows are calculated in a similar manner. Location Abba Sq Ft ; Fraction Sunlit Inside Vbketian Blind Shading Factor Solar Heat Gain Btu/sq ft Radia tion Heat Gain* Btu/hr Normal' Heat Trans. Btu/hr Total Heat . Gain Btu/hr South Windows . South Doors Eastb (Glass)-doors | North Windows ' Total. 60 .35 18 35 30 0.902 1.00 0.74 42' . 42 15 15 V 1680 1470 270 .450 940 2620 550 2020 550 1 820 .470 920 6380 * See Equation 10 and the note under caption of Table 18. b Doors are 1 glass. Calculate sky. radiation for glass portion and normal transmission for entire door assuming U -- 1.04 for wood portion as well as glass. % ' In some jobs it would be desirable to increase (or decrease) the instantaneous radir. 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. Heat Gain from Ventilation and Infiltration: Since the necessary ventilation rate 1275 cfm is greater than one air change per hour, it will be satisfactory for determining.the ventilation, component of the heat gain. : 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